RBSE Solutions Class 12 Maths Chapter 9 समाकलन Exercise 9.4

Question 1. निम्नलिखित फलनों का x के सापेक्ष समाकलन कीजिए \( \int \frac { 1 }{ 16-9x^2 } dx \)
Answer:Let's integrate the given function with respect to \( x \). First, rewrite the denominator to match the standard form for integration. We can take out 9 from the denominator. \( \int \frac { 1 }{ 16-9x^2 } dx = \int \frac { 1 }{ 9( \frac { 16 }{ 9 } - x^2 ) } dx \) \( = \frac { 1 }{ 9 } \int \frac { 1 }{ ( \frac { 4 }{ 3 } )^2 - x^2 } dx \) We use the standard integration formula: \( \int \frac { 1 }{ a^2-x^2 } dx = \frac { 1 }{ 2a } \log \left| \frac { a+x }{ a-x } \right| + C \) Here, \( a = \frac { 4 }{ 3 } \). \( = \frac { 1 }{ 9 } \times \frac { 1 }{ 2 \times \frac { 4 }{ 3 } } \log \left| \frac { \frac { 4 }{ 3 } + x }{ \frac { 4 }{ 3 } - x } \right| + C \) \( = \frac { 1 }{ 9 } \times \frac { 1 }{ \frac { 8 }{ 3 } } \log \left| \frac { \frac { 4+3x }{ 3 } }{ \frac { 4-3x }{ 3 } } \right| + C \) \( = \frac { 1 }{ 9 } \times \frac { 3 }{ 8 } \log \left| \frac { 4+3x }{ 4-3x } \right| + C \) \( = \frac { 1 }{ 24 } \log \left| \frac { 4+3x }{ 4-3x } \right| + C \)
In simple words: We changed the bottom part of the fraction to a common form. Then, we used a special rule for integrals to find the answer. The goal is to make the problem fit a known integration pattern.

🎯 Exam Tip: Always look for common integration formulas after simplifying the expression. Remember to handle constants correctly and the absolute value for logarithm terms.

Question 2. \( \int \frac { 1 }{ x^2-36 } dx \)
Answer:We need to integrate the given function. \( \int \frac { 1 }{ x^2-36 } dx \) First, we can rewrite \( 36 \) as \( 6^2 \). This helps us see a standard integration form. \( = \int \frac { 1 }{ x^2-6^2 } dx \) We use the standard integration formula: \( \int \frac { 1 }{ x^2-a^2 } dx = \frac { 1 }{ 2a } \log \left| \frac { x-a }{ x+a } \right| + C \) Here, \( a = 6 \). \( = \frac { 1 }{ 2 \times 6 } \log \left| \frac { x-6 }{ x+6 } \right| + C \) \( = \frac { 1 }{ 12 } \log \left| \frac { x-6 }{ x+6 } \right| + C \)
In simple words: We changed the number 36 into 6 squared. Then, we used a specific integration rule for fractions that look like this to get the final answer.

🎯 Exam Tip: Recognizing expressions like \( x^2-a^2 \) is key for applying standard integration formulas quickly. Always remember the absolute value inside the logarithm.

Question 3. \( \int \frac { 3x }{ (x+1)(x-2) } dx \)
Answer:To integrate this function, we use partial fraction decomposition because the denominator is a product of linear factors. First, we write the fraction as a sum of simpler fractions: \( \frac { 3x }{ (x+1)(x-2) } = \frac { A }{ x+1 } + \frac { B }{ x-2 } \) To find the values of \( A \) and \( B \), we combine the right side: \( 3x = A(x-2) + B(x+1) \) Now, we can find \( A \) and \( B \) by choosing suitable values for \( x \): If we set \( x = 2 \): \( 3(2) = A(2-2) + B(2+1) \) \( 6 = 0 + 3B \) \( \implies B = 2 \) If we set \( x = -1 \): \( 3(-1) = A(-1-2) + B(-1+1) \) \( -3 = A(-3) + 0 \) \( \implies A = 1 \) So, the integral becomes: \( \int \frac { 3x }{ (x+1)(x-2) } dx = \int \left( \frac { 1 }{ x+1 } + \frac { 2 }{ x-2 } \right) dx \) Now, we integrate each term separately: \( = \int \frac { 1 }{ x+1 } dx + \int \frac { 2 }{ x-2 } dx \) \( = \log |x+1| + 2 \log |x-2| + C \)
In simple words: We broke the complicated fraction into two simpler ones. We found the missing numbers for these simple fractions. Then, we integrated each simple fraction using logarithm rules to get the final answer.

🎯 Exam Tip: Partial fraction decomposition is crucial when the integrand is a rational function with a factorable denominator. Remember to solve for A, B, C accurately.

Question 4. \( \int \frac { 3x-2 }{ (x+1)^2(x+3) } dx \)
Answer:To solve this integral, we use partial fraction decomposition because the denominator has repeated and linear factors. We express the fraction as: \( \frac { 3x-2 }{ (x+1)^2(x+3) } = \frac { A }{ x+1 } + \frac { B }{ (x+1)^2 } + \frac { C }{ x+3 } \) Now, we combine the terms on the right side: \( 3x-2 = A(x+1)(x+3) + B(x+3) + C(x+1)^2 \) Expand the right side: \( 3x-2 = A(x^2+4x+3) + B(x+3) + C(x^2+2x+1) \) \( 3x-2 = Ax^2+4Ax+3A + Bx+3B + Cx^2+2Cx+C \) Group terms by powers of \( x \): \( 3x-2 = (A+C)x^2 + (4A+B+2C)x + (3A+3B+C) \) Compare the coefficients of \( x^2 \), \( x \), and the constant term on both sides: 1. Coefficient of \( x^2 \): \( A+C = 0 \)
\( \implies A = -C \) ....(i) 2. Coefficient of \( x \): \( 4A+B+2C = 3 \) ....(ii) 3. Constant term: \( 3A+3B+C = -2 \) ....(iii) Substitute \( A = -C \) into (ii): \( 4(-C)+B+2C = 3 \) \( -4C+B+2C = 3 \) \( \implies B-2C = 3 \) ....(iv) Substitute \( A = -C \) into (iii): \( 3(-C)+3B+C = -2 \) \( -3C+3B+C = -2 \) \( \implies 3B-2C = -2 \) ....(v) Subtract equation (iv) from equation (v): \( (3B-2C) - (B-2C) = -2 - 3 \) \( 2B = -5 \) \( \implies B = -\frac{5}{2} \) Substitute \( B = -\frac{5}{2} \) into equation (iv): \( -\frac{5}{2} - 2C = 3 \) \( -2C = 3 + \frac{5}{2} \) \( -2C = \frac{6+5}{2} \) \( -2C = \frac{11}{2} \) \( \implies C = -\frac{11}{4} \) Substitute \( C = -\frac{11}{4} \) into equation (i): \( A = -(-\frac{11}{4}) \) \( \implies A = \frac{11}{4} \) Now, substitute the values of \( A, B, C \) back into the partial fraction form: \( \int \frac { 3x-2 }{ (x+1)^2(x+3) } dx = \int \left( \frac { \frac{11}{4} }{ x+1 } + \frac { -\frac{5}{2} }{ (x+1)^2 } + \frac { -\frac{11}{4} }{ x+3 } \right) dx \) \( = \frac { 11 }{ 4 } \int \frac { 1 }{ x+1 } dx - \frac { 5 }{ 2 } \int \frac { 1 }{ (x+1)^2 } dx - \frac { 11 }{ 4 } \int \frac { 1 }{ x+3 } dx \) \( = \frac { 11 }{ 4 } \log |x+1| - \frac { 5 }{ 2 } \left( \frac { -1 }{ x+1 } \right) - \frac { 11 }{ 4 } \log |x+3| + C \) \( = \frac { 11 }{ 4 } \log |x+1| + \frac { 5 }{ 2(x+1) } - \frac { 11 }{ 4 } \log |x+3| + C \) Combine the logarithm terms: \( = \frac { 11 }{ 4 } \log \left| \frac { x+1 }{ x+3 } \right| + \frac { 5 }{ 2(x+1) } + C \)
In simple words: We broke the complex fraction into three simpler fractions because the bottom part had a repeated factor. We found the correct numbers for these fractions by comparing terms. Finally, we integrated each simple part separately to get the result.

🎯 Exam Tip: When dealing with repeated factors in the denominator for partial fractions, remember to include terms for each power of the repeated factor, from 1 up to the highest power.

Question 5. \( \int \frac { x^2 }{ (x+1)(x-2)(x-3) } dx \)
Answer:To integrate this rational function, we use partial fraction decomposition since the denominator is a product of distinct linear factors. We express the fraction as: \( \frac { x^2 }{ (x+1)(x-2)(x-3) } = \frac { A }{ x+1 } + \frac { B }{ x-2 } + \frac { C }{ x-3 } \) Multiply both sides by \( (x+1)(x-2)(x-3) \) to clear the denominators: \( x^2 = A(x-2)(x-3) + B(x+1)(x-3) + C(x+1)(x-2) \) To find the constants \( A, B, C \), we can substitute specific values of \( x \): Set \( x = -1 \): \( (-1)^2 = A(-1-2)(-1-3) + B(0) + C(0) \) \( 1 = A(-3)(-4) \) \( 1 = 12A \) \( \implies A = \frac{1}{12} \) Set \( x = 2 \): \( (2)^2 = A(0) + B(2+1)(2-3) + C(0) \) \( 4 = B(3)(-1) \) \( 4 = -3B \) \( \implies B = -\frac{4}{3} \) Set \( x = 3 \): \( (3)^2 = A(0) + B(0) + C(3+1)(3-2) \) \( 9 = C(4)(1) \) \( 9 = 4C \) \( \implies C = \frac{9}{4} \) Now, substitute the values of \( A, B, C \) back into the partial fraction form: \( \int \frac { x^2 }{ (x+1)(x-2)(x-3) } dx = \int \left( \frac { \frac{1}{12} }{ x+1 } + \frac { -\frac{4}{3} }{ x-2 } + \frac { \frac{9}{4} }{ x-3 } \right) dx \) Integrate each term: \( = \frac { 1 }{ 12 } \int \frac { 1 }{ x+1 } dx - \frac { 4 }{ 3 } \int \frac { 1 }{ x-2 } dx + \frac { 9 }{ 4 } \int \frac { 1 }{ x-3 } dx \) \( = \frac { 1 }{ 12 } \log |x+1| - \frac { 4 }{ 3 } \log |x-2| + \frac { 9 }{ 4 } \log |x-3| + C \)
In simple words: We split the complex fraction into three simpler ones. We found the numbers for these parts by picking easy values for \( x \). Then, we integrated each simple fraction using logarithm rules to find the answer.

🎯 Exam Tip: For distinct linear factors in the denominator, the 'cover-up method' (substituting roots of the denominator factors) is usually the quickest way to find the constants \( A, B, C \).

Question 6. \( \int \frac { x^2 }{ x^4-x^2-12 } dx \)
Answer:To integrate this, we can use a substitution to simplify the denominator. Let \( x^2 = y \). Then the fraction becomes \( \frac { y }{ y^2-y-12 } \). First, factorize the quadratic in the denominator: \( y^2-y-12 = (y-4)(y+3) \) So, we have \( \frac { y }{ (y-4)(y+3) } \). We use partial fraction decomposition for this expression. \( \frac { y }{ (y-4)(y+3) } = \frac { A }{ y-4 } + \frac { B }{ y+3 } \) Multiply both sides by \( (y-4)(y+3) \): \( y = A(y+3) + B(y-4) \) To find \( A \) and \( B \): Set \( y = 4 \): \( 4 = A(4+3) + B(0) \) \( 4 = 7A \) \( \implies A = \frac{4}{7} \) Set \( y = -3 \): \( -3 = A(0) + B(-3-4) \) \( -3 = -7B \) \( \implies B = \frac{3}{7} \) So, \( \frac { x^2 }{ x^4-x^2-12 } = \frac { \frac{4}{7} }{ x^2-4 } + \frac { \frac{3}{7} }{ x^2+3 } \) Now, we integrate the original expression: \( \int \frac { x^2 }{ x^4-x^2-12 } dx = \int \left( \frac { \frac{4}{7} }{ x^2-4 } + \frac { \frac{3}{7} }{ x^2+3 } \right) dx \) \( = \frac { 4 }{ 7 } \int \frac { 1 }{ x^2-2^2 } dx + \frac { 3 }{ 7 } \int \frac { 1 }{ x^2+(\sqrt{3})^2 } dx \) Use standard integration formulas: \( \int \frac { 1 }{ x^2-a^2 } dx = \frac { 1 }{ 2a } \log \left| \frac { x-a }{ x+a } \right| + C \) \( \int \frac { 1 }{ x^2+a^2 } dx = \frac { 1 }{ a } \tan^{-1} \left( \frac { x }{ a } \right) + C \) \( = \frac { 4 }{ 7 } \left( \frac { 1 }{ 2 \times 2 } \log \left| \frac { x-2 }{ x+2 } \right| \right) + \frac { 3 }{ 7 } \left( \frac { 1 }{ \sqrt{3} } \tan^{-1} \left( \frac { x }{ \sqrt{3} } \right) \right) + C \) \( = \frac { 4 }{ 7 } \times \frac { 1 }{ 4 } \log \left| \frac { x-2 }{ x+2 } \right| + \frac { 3 }{ 7\sqrt{3} } \tan^{-1} \left( \frac { x }{ \sqrt{3} } \right) + C \) \( = \frac { 1 }{ 7 } \log \left| \frac { x-2 }{ x+2 } \right| + \frac { \sqrt{3} }{ 7 } \tan^{-1} \left( \frac { x }{ \sqrt{3} } \right) + C \)
In simple words: We first replaced \( x^2 \) with \( y \) to simplify the fraction, then broke it into smaller parts using partial fractions. After finding the numbers for these parts, we put \( x^2 \) back in and used known integration rules to solve each piece. This method helps to handle more complex fractions.

🎯 Exam Tip: For integrands involving only even powers of \( x \), like \( x^2 \) and \( x^4 \), a substitution like \( y = x^2 \) often helps simplify the partial fraction decomposition step before final integration.

Question 7. \( \int \frac { 1 }{ x^3-x^2-x+1 } dx \)
Answer:To integrate this function, we first factorize the denominator. \( x^3-x^2-x+1 = x^2(x-1) - 1(x-1) \) \( = (x^2-1)(x-1) \) \( = (x-1)(x+1)(x-1) \) \( = (x-1)^2(x+1) \) Now, we use partial fraction decomposition for the integrand: \( \frac { 1 }{ (x-1)^2(x+1) } = \frac { A }{ x-1 } + \frac { B }{ (x-1)^2 } + \frac { C }{ x+1 } \) Multiply both sides by \( (x-1)^2(x+1) \): \( 1 = A(x-1)(x+1) + B(x+1) + C(x-1)^2 \) Expand the right side: \( 1 = A(x^2-1) + B(x+1) + C(x^2-2x+1) \) \( 1 = Ax^2-A + Bx+B + Cx^2-2Cx+C \) Group terms by powers of \( x \): \( 1 = (A+C)x^2 + (B-2C)x + (-A+B+C) \) Compare coefficients: 1. Coefficient of \( x^2 \): \( A+C = 0 \)
\( \implies A = -C \) 2. Coefficient of \( x \): \( B-2C = 0 \)
\( \implies B = 2C \) 3. Constant term: \( -A+B+C = 1 \) Substitute \( A = -C \) and \( B = 2C \) into the constant term equation: \( -(-C) + (2C) + C = 1 \) \( C+2C+C = 1 \) \( 4C = 1 \) \( \implies C = \frac{1}{4} \) Now, find \( A \) and \( B \): \( A = -C = -\frac{1}{4} \) \( B = 2C = 2 \times \frac{1}{4} = \frac{1}{2} \) Substitute these values back into the partial fraction form: \( \int \frac { 1 }{ (x-1)^2(x+1) } dx = \int \left( \frac { -\frac{1}{4} }{ x-1 } + \frac { \frac{1}{2} }{ (x-1)^2 } + \frac { \frac{1}{4} }{ x+1 } \right) dx \) Integrate each term: \( = -\frac { 1 }{ 4 } \int \frac { 1 }{ x-1 } dx + \frac { 1 }{ 2 } \int \frac { 1 }{ (x-1)^2 } dx + \frac { 1 }{ 4 } \int \frac { 1 }{ x+1 } dx \) \( = -\frac { 1 }{ 4 } \log |x-1| + \frac { 1 }{ 2 } \left( \frac { -1 }{ x-1 } \right) + \frac { 1 }{ 4 } \log |x+1| + C \) \( = -\frac { 1 }{ 4 } \log |x-1| - \frac { 1 }{ 2(x-1) } + \frac { 1 }{ 4 } \log |x+1| + C \) We can combine the logarithm terms: \( = \frac { 1 }{ 4 } (\log |x+1| - \log |x-1|) - \frac { 1 }{ 2(x-1) } + C \) \( = \frac { 1 }{ 4 } \log \left| \frac { x+1 }{ x-1 } \right| - \frac { 1 }{ 2(x-1) } + C \)
In simple words: First, we cleaned up the bottom part of the fraction by factoring it. Since there was a repeated part, we broke the fraction into three simpler pieces. After finding the numbers for these pieces, we integrated each one to get the final answer.

🎯 Exam Tip: Always factorize the denominator completely before setting up partial fractions. Remember that a term like \( (x-a)^2 \) requires two partial fractions: \( \frac{A}{x-a} \) and \( \frac{B}{(x-a)^2} \).

Question 8. \( \int \frac { x^2 }{ (x+1)(x-2) } dx \)
Answer:In this integral, the degree of the numerator (2) is equal to the degree of the denominator (product of \( (x+1)(x-2) \) which is \( x^2-x-2 \)). When the degrees are equal or the numerator's degree is higher, we must perform polynomial long division first. \( \frac { x^2 }{ x^2-x-2 } = 1 + \frac { x+2 }{ x^2-x-2 } \) So the integral becomes: \( \int \left( 1 + \frac { x+2 }{ (x+1)(x-2) } \right) dx = \int 1 dx + \int \frac { x+2 }{ (x+1)(x-2) } dx \) For the second part, we use partial fraction decomposition: \( \frac { x+2 }{ (x+1)(x-2) } = \frac { A }{ x+1 } + \frac { B }{ x-2 } \) Multiply both sides by \( (x+1)(x-2) \): \( x+2 = A(x-2) + B(x+1) \) To find \( A \) and \( B \): Set \( x = 2 \): \( 2+2 = A(0) + B(2+1) \) \( 4 = 3B \) \( \implies B = \frac{4}{3} \) Set \( x = -1 \): \( -1+2 = A(-1-2) + B(0) \) \( 1 = -3A \) \( \implies A = -\frac{1}{3} \) Now substitute \( A \) and \( B \) back into the integral: \( \int \frac { x+2 }{ (x+1)(x-2) } dx = \int \left( \frac { -\frac{1}{3} }{ x+1 } + \frac { \frac{4}{3} }{ x-2 } \right) dx \) \( = -\frac { 1 }{ 3 } \int \frac { 1 }{ x+1 } dx + \frac { 4 }{ 3 } \int \frac { 1 }{ x-2 } dx \) \( = -\frac { 1 }{ 3 } \log |x+1| + \frac { 4 }{ 3 } \log |x-2| + C' \) Combine this with the first part of the integral \( \int 1 dx = x \): The full integral is: \( = x - \frac { 1 }{ 3 } \log |x+1| + \frac { 4 }{ 3 } \log |x-2| + C \) This can also be written as: \( = x + \frac { 1 }{ 3 } (4 \log |x-2| - \log |x+1|) + C \) \( = x + \frac { 1 }{ 3 } \log \left| \frac { (x-2)^4 }{ x+1 } \right| + C \)
In simple words: Since the top and bottom of the fraction had the same highest power, we first divided them like regular numbers. This gave us a whole number part and a new fraction. Then, we broke the new fraction into simpler pieces and found the correct numbers for them. Finally, we added up the integrals of all these simple parts to get the total answer.

🎯 Exam Tip: Always perform polynomial long division if the degree of the numerator is greater than or equal to the degree of the denominator before attempting partial fraction decomposition.

Question 9. \( \int \frac { x^2 }{ (x^2+a^2)(x^2+b^2) } dx \)
Answer:To integrate this function, we can simplify it by using a substitution for \( x^2 \). Let \( x^2 = y \). The fraction becomes \( \frac { y }{ (y+a^2)(y+b^2) } \). Now, we use partial fraction decomposition: \( \frac { y }{ (y+a^2)(y+b^2) } = \frac { A }{ y+a^2 } + \frac { B }{ y+b^2 } \) Multiply both sides by \( (y+a^2)(y+b^2) \): \( y = A(y+b^2) + B(y+a^2) \) To find \( A \) and \( B \): Set \( y = -a^2 \): \( -a^2 = A(-a^2+b^2) + B(0) \) \( -a^2 = A(b^2-a^2) \) \( \implies A = \frac { -a^2 }{ b^2-a^2 } = \frac { a^2 }{ a^2-b^2 } \) Set \( y = -b^2 \): \( -b^2 = A(0) + B(-b^2+a^2) \) \( -b^2 = B(a^2-b^2) \) \( \implies B = \frac { -b^2 }{ a^2-b^2 } \) Now, substitute these back into the integral, replacing \( y \) with \( x^2 \): \( \int \frac { x^2 }{ (x^2+a^2)(x^2+b^2) } dx = \int \left( \frac { \frac{a^2}{a^2-b^2} }{ x^2+a^2 } + \frac { \frac{-b^2}{a^2-b^2} }{ x^2+b^2 } \right) dx \) \( = \frac { a^2 }{ a^2-b^2 } \int \frac { 1 }{ x^2+a^2 } dx - \frac { b^2 }{ a^2-b^2 } \int \frac { 1 }{ x^2+b^2 } dx \) Use the standard integral formula: \( \int \frac { 1 }{ x^2+k^2 } dx = \frac { 1 }{ k } \tan^{-1} \left( \frac { x }{ k } \right) + C \) \( = \frac { a^2 }{ a^2-b^2 } \left( \frac { 1 }{ a } \tan^{-1} \left( \frac { x }{ a } \right) \right) - \frac { b^2 }{ a^2-b^2 } \left( \frac { 1 }{ b } \tan^{-1} \left( \frac { x }{ b } ) \right) + C \) \( = \frac { a }{ a^2-b^2 } \tan^{-1} \left( \frac { x }{ a } \right) - \frac { b }{ a^2-b^2 } \tan^{-1} \left( \frac { x }{ b } \right) + C \) Factor out \( \frac { 1 }{ a^2-b^2 } \): \( = \frac { 1 }{ a^2-b^2 } \left[ a \tan^{-1} \left( \frac { x }{ a } \right) - b \tan^{-1} \left( \frac { x }{ b } \right) \right] + C \)
In simple words: We used a trick by replacing \( x^2 \) with \( y \) to make the fraction simpler. Then, we broke this new fraction into two parts and found the numbers for each part. After putting \( x^2 \) back, we used a special rule for inverse tangent functions to solve each part. This makes integrating fractions with squared terms easier.

🎯 Exam Tip: When the integrand contains only even powers of \( x \) in terms like \( (x^2+a^2)(x^2+b^2) \), substituting \( y=x^2 \) for partial fractions is a common and effective technique, but remember to integrate with respect to \( x \) in the end.

Question 10. \( \int \frac { x+1 }{ x^3+x^2-6x } dx \)
Answer:To solve this integral, we first need to factorize the denominator completely. \( x^3+x^2-6x = x(x^2+x-6) \) Now, factorize the quadratic term \( x^2+x-6 \): \( x^2+x-6 = (x+3)(x-2) \) So, the denominator is \( x(x+3)(x-2) \). Now, we use partial fraction decomposition for the integrand: \( \frac { x+1 }{ x(x+3)(x-2) } = \frac { A }{ x } + \frac { B }{ x+3 } + \frac { C }{ x-2 } \) Multiply both sides by \( x(x+3)(x-2) \): \( x+1 = A(x+3)(x-2) + Bx(x-2) + Cx(x+3) \) To find the constants \( A, B, C \), we can substitute specific values of \( x \): Set \( x = 0 \): \( 0+1 = A(0+3)(0-2) + B(0) + C(0) \) \( 1 = A(3)(-2) \) \( 1 = -6A \) \( \implies A = -\frac{1}{6} \) Set \( x = 2 \): \( 2+1 = A(0) + B(0) + C(2)(2+3) \) \( 3 = C(2)(5) \) \( 3 = 10C \) \( \implies C = \frac{3}{10} \) Set \( x = -3 \): \( -3+1 = A(0) + B(-3)(-3-2) + C(0) \) \( -2 = B(-3)(-5) \) \( -2 = 15B \) \( \implies B = -\frac{2}{15} \) Now, substitute the values of \( A, B, C \) back into the partial fraction form: \( \int \frac { x+1 }{ x(x+3)(x-2) } dx = \int \left( \frac { -\frac{1}{6} }{ x } + \frac { -\frac{2}{15} }{ x+3 } + \frac { \frac{3}{10} }{ x-2 } \right) dx \) Integrate each term: \( = -\frac { 1 }{ 6 } \int \frac { 1 }{ x } dx - \frac { 2 }{ 15 } \int \frac { 1 }{ x+3 } dx + \frac { 3 }{ 10 } \int \frac { 1 }{ x-2 } dx \) \( = -\frac { 1 }{ 6 } \log |x| - \frac { 2 }{ 15 } \log |x+3| + \frac { 3 }{ 10 } \log |x-2| + C \)
In simple words: We first broke down the bottom part of the fraction into simpler multiplication terms. Then, we used a method to split the whole fraction into three even simpler fractions. After finding the correct numbers for these parts, we integrated each simple fraction using logarithm rules to get our final answer.

🎯 Exam Tip: Always factorize the denominator completely before applying partial fraction decomposition. This includes taking out common factors like 'x' if present.

Question 11. \( \int \frac { x^2-8x+4 }{ x^3-4x } dx \)
Answer:To solve this integral, we begin by factoring the denominator. \( x^3-4x = x(x^2-4) \) We can further factorize \( x^2-4 \) using the difference of squares formula \( a^2-b^2=(a-b)(a+b) \): \( x^2-4 = (x-2)(x+2) \) So, the denominator is \( x(x-2)(x+2) \). Now, we use partial fraction decomposition for the integrand: \( \frac { x^2-8x+4 }{ x(x-2)(x+2) } = \frac { A }{ x } + \frac { B }{ x-2 } + \frac { C }{ x+2 } \) Multiply both sides by \( x(x-2)(x+2) \): \( x^2-8x+4 = A(x-2)(x+2) + Bx(x+2) + Cx(x-2) \) To find the constants \( A, B, C \), we substitute specific values of \( x \): Set \( x = 0 \): \( (0)^2-8(0)+4 = A(0-2)(0+2) + B(0) + C(0) \) \( 4 = A(-2)(2) \) \( 4 = -4A \) \( \implies A = -1 \) Set \( x = 2 \): \( (2)^2-8(2)+4 = A(0) + B(2)(2+2) + C(0) \) \( 4-16+4 = B(2)(4) \) \( -8 = 8B \) \( \implies B = -1 \) Set \( x = -2 \): \( (-2)^2-8(-2)+4 = A(0) + B(0) + C(-2)(-2-2) \) \( 4+16+4 = C(-2)(-4) \) \( 24 = 8C \) \( \implies C = 3 \) Now, substitute the values of \( A, B, C \) back into the partial fraction form: \( \int \frac { x^2-8x+4 }{ x(x-2)(x+2) } dx = \int \left( \frac { -1 }{ x } + \frac { -1 }{ x-2 } + \frac { 3 }{ x+2 } \right) dx \) Integrate each term: \( = -\int \frac { 1 }{ x } dx - \int \frac { 1 }{ x-2 } dx + 3\int \frac { 1 }{ x+2 } dx \) \( = -\log |x| - \log |x-2| + 3\log |x+2| + C \)
In simple words: We first simplified the bottom part of the fraction by finding its factors. Then, we broke the whole fraction into three simpler pieces. After finding the correct numbers for these pieces, we integrated each simple fraction using logarithm rules to find our final answer.

🎯 Exam Tip: Always remember to factorize the denominator completely, using methods like taking common factors and difference of squares, before proceeding with partial fractions.

Question 12. \( \int \frac { 1 }{ (x-1)^2(x+2) } dx \)
Answer:To integrate this function, we use partial fraction decomposition because the denominator has a repeated linear factor. We express the fraction as: \( \frac { 1 }{ (x-1)^2(x+2) } = \frac { A }{ x-1 } + \frac { B }{ (x-1)^2 } + \frac { C }{ x+2 } \) Multiply both sides by \( (x-1)^2(x+2) \) to clear the denominators: \( 1 = A(x-1)(x+2) + B(x+2) + C(x-1)^2 \) Expand the right side: \( 1 = A(x^2+x-2) + B(x+2) + C(x^2-2x+1) \) \( 1 = Ax^2+Ax-2A + Bx+2B + Cx^2-2Cx+C \) Group terms by powers of \( x \): \( 1 = (A+C)x^2 + (A+B-2C)x + (-2A+2B+C) \) Compare the coefficients of \( x^2 \), \( x \), and the constant term: 1. Coefficient of \( x^2 \): \( A+C = 0 \)
\( \implies A = -C \) 2. Coefficient of \( x \): \( A+B-2C = 0 \) 3. Constant term: \( -2A+2B+C = 1 \) Substitute \( A = -C \) into the second equation: \( -C+B-2C = 0 \) \( B-3C = 0 \)
\( \implies B = 3C \) Substitute \( A = -C \) and \( B = 3C \) into the third equation: \( -2(-C) + 2(3C) + C = 1 \) \( 2C+6C+C = 1 \) \( 9C = 1 \) \( \implies C = \frac{1}{9} \) Now, find \( A \) and \( B \): \( A = -C = -\frac{1}{9} \) \( B = 3C = 3 \times \frac{1}{9} = \frac{1}{3} \) Substitute these values back into the partial fraction form: \( \int \frac { 1 }{ (x-1)^2(x+2) } dx = \int \left( \frac { -\frac{1}{9} }{ x-1 } + \frac { \frac{1}{3} }{ (x-1)^2 } + \frac { \frac{1}{9} }{ x+2 } \right) dx \) Integrate each term: \( = -\frac { 1 }{ 9 } \int \frac { 1 }{ x-1 } dx + \frac { 1 }{ 3 } \int \frac { 1 }{ (x-1)^2 } dx + \frac { 1 }{ 9 } \int \frac { 1 }{ x+2 } dx \) \( = -\frac { 1 }{ 9 } \log |x-1| + \frac { 1 }{ 3 } \left( \frac { -1 }{ x-1 } \right) + \frac { 1 }{ 9 } \log |x+2| + C \) \( = -\frac { 1 }{ 9 } \log |x-1| - \frac { 1 }{ 3(x-1) } + \frac { 1 }{ 9 } \log |x+2| + C \) Combine the logarithm terms: \( = \frac { 1 }{ 9 } (\log |x+2| - \log |x-1|) - \frac { 1 }{ 3(x-1) } + C \) \( = \frac { 1 }{ 9 } \log \left| \frac { x+2 }{ x-1 } \right| - \frac { 1 }{ 3(x-1) } + C \)
In simple words: We separated the fraction into three simpler parts because the bottom had a repeated factor. By comparing the terms, we found the correct numbers for each part. Then, we integrated each simple fraction using logarithm and power rules to get the final answer.

🎯 Exam Tip: For repeated factors in the denominator, remember that for a factor \( (x-a)^n \), you need \( n \) partial fractions: \( \frac{A_1}{x-a} + \frac{A_2}{(x-a)^2} + \dots + \frac{A_n}{(x-a)^n} \).

Question 13. \( \int \frac { 1-3x }{ 1+x+x^2+x^3 } dx \)
Answer:To integrate this function, we first factorize the denominator completely. \( 1+x+x^2+x^3 = (1+x) + x^2(1+x) \) \( = (1+x)(1+x^2) \) Now, we use partial fraction decomposition for the integrand. Since \( x^2+1 \) is an irreducible quadratic factor, its numerator must be a linear term \( Bx+C \). \( \frac { 1-3x }{ (1+x)(1+x^2) } = \frac { A }{ 1+x } + \frac { Bx+C }{ 1+x^2 } \) Multiply both sides by \( (1+x)(1+x^2) \): \( 1-3x = A(1+x^2) + (Bx+C)(1+x) \) Expand the right side: \( 1-3x = A+Ax^2 + Bx+Bx^2 + C+Cx \) Group terms by powers of \( x \): \( 1-3x = (A+B)x^2 + (B+C)x + (A+C) \) Compare coefficients: 1. Coefficient of \( x^2 \): \( A+B = 0 \)
\( \implies B = -A \) 2. Coefficient of \( x \): \( B+C = -3 \) 3. Constant term: \( A+C = 1 \) Substitute \( B = -A \) into the second equation: \( -A+C = -3 \) Now we have two equations with \( A \) and \( C \): \( A+C = 1 \) (from constant term) \( -A+C = -3 \) (from coefficient of \( x \)) Add these two equations: \( (A+C) + (-A+C) = 1 + (-3) \) \( 2C = -2 \) \( \implies C = -1 \) Substitute \( C = -1 \) into \( A+C=1 \): \( A-1 = 1 \) \( \implies A = 2 \) Now find \( B \): \( B = -A = -2 \) Substitute these values back into the partial fraction form: \( \int \frac { 1-3x }{ (1+x)(1+x^2) } dx = \int \left( \frac { 2 }{ 1+x } + \frac { -2x-1 }{ 1+x^2 } \right) dx \) We can split the second term: \( = \int \frac { 2 }{ 1+x } dx - \int \frac { 2x }{ 1+x^2 } dx - \int \frac { 1 }{ 1+x^2 } dx \) Integrate each term: 1. \( \int \frac { 2 }{ 1+x } dx = 2 \log |1+x| \) 2. For \( \int \frac { 2x }{ 1+x^2 } dx \), let \( u = 1+x^2 \implies du = 2x dx \). So, this integral is \( \int \frac { du }{ u } = \log |u| = \log |1+x^2| \). 3. \( \int \frac { 1 }{ 1+x^2 } dx = \tan^{-1}(x) \) Combining these, the full integral is: \( = 2 \log |1+x| - \log |1+x^2| - \tan^{-1}(x) + C \) Using logarithm properties, \( 2 \log |1+x| = \log |(1+x)^2| \): \( = \log \left| \frac { (1+x)^2 }{ 1+x^2 } \right| - \tan^{-1}(x) + C \)
In simple words: First, we broke the bottom part of the fraction into two factors. One of these was a quadratic that could not be factored further. We then split the main fraction into two simpler parts, one for each factor, and found their correct numerical values. Finally, we integrated each part, using logarithm rules for some parts and the inverse tangent rule for others, to reach the answer.

🎯 Exam Tip: When setting up partial fractions, if a quadratic factor \( (ax^2+bx+c) \) cannot be further factored into linear terms, its corresponding numerator in the partial fraction must be of the form \( Dx+E \).

Question 14. \( \int \frac { 1+x^2 }{ x(x-1)(x+1) } dx \)
Answer:To integrate this function, we use partial fraction decomposition. The denominator is already factored into distinct linear terms. First, we rewrite the denominator: \( x(x-1)(x+1) = x(x^2-1) \). Now, we express the integrand as a sum of partial fractions: \( \frac { 1+x^2 }{ x(x-1)(x+1) } = \frac { A }{ x } + \frac { B }{ x-1 } + \frac { C }{ x+1 } \) Multiply both sides by \( x(x-1)(x+1) \) to clear the denominators: \( 1+x^2 = A(x-1)(x+1) + Bx(x+1) + Cx(x-1) \) Expand the right side: \( 1+x^2 = A(x^2-1) + B(x^2+x) + C(x^2-x) \) \( 1+x^2 = Ax^2-A + Bx^2+Bx + Cx^2-Cx \) Group terms by powers of \( x \): \( 1+x^2 = (A+B+C)x^2 + (B-C)x - A \) Compare the coefficients of \( x^2 \), \( x \), and the constant term on both sides: 1. Coefficient of \( x^2 \): \( A+B+C = 1 \) 2. Coefficient of \( x \): \( B-C = 0 \)
\( \implies B = C \) 3. Constant term: \( -A = 1 \)
\( \implies A = -1 \) Substitute \( A = -1 \) and \( B = C \) into the first equation: \( -1+C+C = 1 \) \( -1+2C = 1 \) \( 2C = 2 \) \( \implies C = 1 \) Since \( B=C \), then \( B = 1 \). So, the values are \( A=-1, B=1, C=1 \). Now, substitute these values back into the partial fraction form: \( \int \frac { 1+x^2 }{ x(x-1)(x+1) } dx = \int \left( \frac { -1 }{ x } + \frac { 1 }{ x-1 } + \frac { 1 }{ x+1 } \right) dx \) Integrate each term: \( = -\int \frac { 1 }{ x } dx + \int \frac { 1 }{ x-1 } dx + \int \frac { 1 }{ x+1 } dx \) \( = -\log |x| + \log |x-1| + \log |x+1| + C \) Combine the logarithm terms using the property \( \log a + \log b = \log (ab) \) and \( -\log a = \log (1/a) \): \( = \log |x-1| + \log |x+1| - \log |x| + C \) \( = \log |(x-1)(x+1)| - \log |x| + C \) \( = \log |x^2-1| - \log |x| + C \) \( = \log \left| \frac { x^2-1 }{ x } \right| + C \)
In simple words: We separated the fraction into three simpler pieces because its bottom part was made of three different single-term factors. We found the correct numbers for these pieces by comparing terms in the expanded equation. Then, we integrated each simple fraction using logarithm rules to get the final answer.

🎯 Exam Tip: Always solve for the constants \( A, B, C \) systematically, either by substituting roots of the denominator or by comparing coefficients. Double-check your algebra to avoid errors.

Question 15. \( \int \frac { x^2+5x+3 }{ x^2+3x+2 } dx \)
Answer:In this integral, the degree of the numerator (2) is equal to the degree of the denominator (2). When the degrees are equal or the numerator's degree is higher, we must perform polynomial long division first. \( \frac { x^2+5x+3 }{ x^2+3x+2 } \) Divide \( x^2+5x+3 \) by \( x^2+3x+2 \): The quotient is 1 and the remainder is \( (x^2+5x+3) - 1(x^2+3x+2) = x^2+5x+3-x^2-3x-2 = 2x+1 \). So, \( \frac { x^2+5x+3 }{ x^2+3x+2 } = 1 + \frac { 2x+1 }{ x^2+3x+2 } \) Now, we factorize the denominator of the new fraction: \( x^2+3x+2 = (x+1)(x+2) \) So the integral becomes: \( \int \left( 1 + \frac { 2x+1 }{ (x+1)(x+2) } \right) dx = \int 1 dx + \int \frac { 2x+1 }{ (x+1)(x+2) } dx \) For the second part, we use partial fraction decomposition: \( \frac { 2x+1 }{ (x+1)(x+2) } = \frac { A }{ x+1 } + \frac { B }{ x+2 } \) Multiply both sides by \( (x+1)(x+2) \): \( 2x+1 = A(x+2) + B(x+1) \) To find \( A \) and \( B \): Set \( x = -1 \): \( 2(-1)+1 = A(-1+2) + B(0) \) \( -2+1 = A(1) \) \( -1 = A \) Set \( x = -2 \): \( 2(-2)+1 = A(0) + B(-2+1) \) \( -4+1 = B(-1) \) \( -3 = -B \) \( \implies B = 3 \) Now substitute \( A \) and \( B \) back into the integral for the fractional part: \( \int \frac { 2x+1 }{ (x+1)(x+2) } dx = \int \left( \frac { -1 }{ x+1 } + \frac { 3 }{ x+2 } \right) dx \) \( = -\int \frac { 1 }{ x+1 } dx + 3\int \frac { 1 }{ x+2 } dx \) \( = -\log |x+1| + 3\log |x+2| + C' \) Combine this with the first part of the integral \( \int 1 dx = x \): The full integral is: \( = x - \log |x+1| + 3\log |x+2| + C \)
In simple words: Since the top and bottom of the fraction had the same highest power, we first divided them. This gave us a whole number plus a new fraction. We then broke this new fraction into simpler parts. After finding the correct numbers for these parts, we added the integrals of all these simple parts, including the whole number part, to get the total answer.

🎯 Exam Tip: Remember to always perform polynomial long division first if the degree of the numerator is greater than or equal to the degree of the denominator in rational function integration problems.

Question 15. \( \int \frac { x^{2} + 5x + 3 }{ x^{2} + 3x + 2 } dx \)
Answer:First, we divide the numerator by the denominator because the degree of the numerator is not less than the degree of the denominator. \[ \frac { x^{2} + 5x + 3 }{ x^{2} + 3x + 2 } = 1 + \frac { 2x + 1 }{ x^{2} + 3x + 2 } \] Now, we use partial fractions for the term \( \frac { 2x + 1 }{ x^{2} + 3x + 2 } \). The denominator can be factored as \( x^{2} + 3x + 2 = (x+1)(x+2) \). Let \( \frac { 2x + 1 }{ (x+1)(x+2) } = \frac { A }{ x+1 } + \frac { B }{ x+2 } \)
\( \implies \) \( 2x + 1 = A(x+2) + B(x+1) \) To find A, set \( x = -1 \): \( 2(-1) + 1 = A(-1+2) + B(-1+1) \)
\( \implies \) \( -2 + 1 = A(1) + B(0) \)
\( \implies \) \( -1 = A \) To find B, set \( x = -2 \): \( 2(-2) + 1 = A(-2+2) + B(-2+1) \)
\( \implies \) \( -4 + 1 = A(0) + B(-1) \)
\( \implies \) \( -3 = -B \)
\( \implies \) \( B = 3 \) So, \( \frac { 2x + 1 }{ (x+1)(x+2) } = \frac { -1 }{ x+1 } + \frac { 3 }{ x+2 } \) Therefore, the integral becomes: \( \int \frac { x^{2} + 5x + 3 }{ x^{2} + 3x + 2 } dx = \int \left( 1 + \frac { -1 }{ x+1 } + \frac { 3 }{ x+2 } \right) dx \) \( = \int 1 dx - \int \frac { 1 }{ x+1 } dx + \int \frac { 3 }{ x+2 } dx \) \( = x - \log|x+1| + 3 \log|x+2| + C \) This process of breaking down complex fractions into simpler ones makes integration easier.
In simple words: First, we divided the top by the bottom of the fraction. Then, we broke the remaining fraction into two simpler parts. Finally, we integrated each part separately to get the answer.

🎯 Exam Tip: When the degree of the numerator is equal to or greater than the denominator, always perform polynomial long division first before applying partial fractions. Remember that \( \int \frac{1}{ax+b} dx = \frac{1}{a} \log|ax+b| + C \).

Question 16. \( \int \frac { x-1 }{ (x+1)(x^{2} + 1) } dx \)
Answer:We use partial fractions to simplify the integrand: Let \( \frac { x-1 }{ (x+1)(x^{2} + 1) } = \frac { A }{ x+1 } + \frac { Bx+C }{ x^{2} + 1 } \)
\( \implies \) \( x-1 = A(x^{2} + 1) + (Bx+C)(x+1) \)
\( \implies \) \( x-1 = Ax^{2} + A + Bx^{2} + Bx + Cx + C \)
\( \implies \) \( x-1 = (A+B)x^{2} + (B+C)x + (A+C) \) Comparing the coefficients of like powers of x: Coefficient of \( x^{2} \): \( A+B = 0 \). (Equation 1) Coefficient of \( x \): \( B+C = 1 \). (Equation 2) Constant term: \( A+C = -1 \). (Equation 3) From Equation 1, \( B = -A \). Substitute B into Equation 2: \( -A+C = 1 \). (Equation 4) Now, we have a system with Equation 3 and Equation 4: \( A+C = -1 \) \( -A+C = 1 \) Adding these two equations: \( (A+C) + (-A+C) = -1 + 1 \)
\( \implies \) \( 2C = 0 \)
\( \implies \) \( C = 0 \) Substitute C=0 into Equation 3: \( A+0 = -1 \implies A = -1 \) Substitute A=-1 into Equation 1: \( -1+B = 0 \implies B = 1 \) So, \( A = -1, B = 1, C = 0 \). Thus, \( \frac { x-1 }{ (x+1)(x^{2} + 1) } = \frac { -1 }{ x+1 } + \frac { 1x+0 }{ x^{2} + 1 } = \frac { -1 }{ x+1 } + \frac { x }{ x^{2} + 1 } \) Now, we integrate: \( \int \frac { x-1 }{ (x+1)(x^{2} + 1) } dx = \int \frac { -1 }{ x+1 } dx + \int \frac { x }{ x^{2} + 1 } dx \) \( = -\int \frac { 1 }{ x+1 } dx + \frac { 1 }{ 2 } \int \frac { 2x }{ x^{2} + 1 } dx \) \( = -\log|x+1| + \frac { 1 }{ 2 } \log|x^{2} + 1| + C \) \( = \log\left| \frac { \sqrt{x^{2}+1} }{ x+1 } \right| + C \) Partial fraction decomposition is key for integrating rational functions.
In simple words: We split the complex fraction into simpler parts. Then, we solved for the missing numbers and integrated each easy part. This gives us the final answer using logarithms.

🎯 Exam Tip: For partial fractions involving quadratic factors like \( x^{2}+1 \), the numerator must be in the form \( Bx+C \). Remember to multiply and compare coefficients carefully to find A, B, and C.

Question 17. \( \int \frac { 1 }{ (1+e^{x})(1-e^{-x}) } dx \)
Answer:First, rewrite \( 1-e^{-x} \) as \( 1-\frac{1}{e^x} = \frac{e^x-1}{e^x} \). So the integrand becomes: \( \frac { 1 }{ (1+e^{x})\left(\frac{e^x-1}{e^x}\right) } = \frac { e^{x} }{ (e^{x}+1)(e^{x}-1) } \) Now, let \( t = e^{x} \).
\( \implies \) \( dt = e^{x} dx \) The integral becomes \( \int \frac { dt }{ (t+1)(t-1) } \) We use partial fractions: Let \( \frac { 1 }{ (t+1)(t-1) } = \frac { A }{ t+1 } + \frac { B }{ t-1 } \)
\( \implies \) \( 1 = A(t-1) + B(t+1) \) To find A, set \( t = -1 \): \( 1 = A(-1-1) + B(-1+1) \)
\( \implies \) \( 1 = -2A \)
\( \implies \) \( A = -\frac{1}{2} \) To find B, set \( t = 1 \): \( 1 = A(1-1) + B(1+1) \)
\( \implies \) \( 1 = 2B \)
\( \implies \) \( B = \frac{1}{2} \) So, \( \int \frac { dt }{ (t+1)(t-1) } = \int \left( \frac { -1/2 }{ t+1 } + \frac { 1/2 }{ t-1 } \right) dt \) \( = -\frac{1}{2} \int \frac { 1 }{ t+1 } dt + \frac{1}{2} \int \frac { 1 }{ t-1 } dt \) \( = -\frac{1}{2} \log|t+1| + \frac{1}{2} \log|t-1| + C \) \( = \frac{1}{2} (\log|t-1| - \log|t+1|) + C \) \( = \frac{1}{2} \log\left|\frac{t-1}{t+1}\right| + C \) Substitute back \( t = e^{x} \): \( = \frac{1}{2} \log\left|\frac{e^{x}-1}{e^{x}+1}\right| + C \) This method of substitution along with partial fractions simplifies complex integrals greatly.
In simple words: We first changed the expression to make it easier to work with. Then, we replaced \( e^x \) with \( t \) and used a trick called partial fractions to break it into two simple parts. After integrating, we put \( e^x \) back to get the final answer.

🎯 Exam Tip: When you see \( e^{-x} \) in the denominator, try to multiply the numerator and denominator by \( e^{x} \) to make the substitution \( t = e^{x} \) more effective. This simplifies the integral for partial fraction decomposition.

Question 19. \( \int \frac { e^{x} }{ (e^{x})^{2} + 5e^{x} + 6 } dx \)
Answer:Let \( t = e^{x} \).
\( \implies \) \( dt = e^{x} dx \) The integral becomes \( \int \frac { dt }{ t^{2} + 5t + 6 } \) First, factor the denominator: \( t^{2} + 5t + 6 = (t+3)(t+2) \) Now, we use partial fractions for \( \frac { 1 }{ (t+3)(t+2) } \): Let \( \frac { 1 }{ (t+3)(t+2) } = \frac { A }{ t+3 } + \frac { B }{ t+2 } \)
\( \implies \) \( 1 = A(t+2) + B(t+3) \) To find A, set \( t = -3 \): \( 1 = A(-3+2) + B(-3+3) \)
\( \implies \) \( 1 = -A \)
\( \implies \) \( A = -1 \) To find B, set \( t = -2 \): \( 1 = A(-2+2) + B(-2+3) \)
\( \implies \) \( 1 = B \) So, \( \int \frac { dt }{ (t+3)(t+2) } = \int \left( \frac { -1 }{ t+3 } + \frac { 1 }{ t+2 } \right) dt \) \( = -\int \frac { 1 }{ t+3 } dt + \int \frac { 1 }{ t+2 } dt \) \( = -\log|t+3| + \log|t+2| + C \) \( = \log\left|\frac{t+2}{t+3}\right| + C \) Substitute back \( t = e^{x} \): \( = \log\left|\frac{e^{x}+2}{e^{x}+3}\right| + C \) Using substitution transforms the integral into a simpler form, making it easier to solve.
In simple words: We replaced \( e^x \) with \( t \), which changed the problem into a simpler fraction. Then, we broke this fraction into two easy parts and integrated them separately. Finally, we put \( e^x \) back to get the answer.

🎯 Exam Tip: When you see \( e^x \) and its square in an integral, try the substitution \( t = e^x \). This often turns the integral into a rational function solvable by partial fractions. Always remember to factor the denominator correctly!

Question 20. \( \int \frac { \sec^{2}x }{ (2+\tan x)(3+\tan x) } dx \)
Answer:Let \( t = \tan x \).
\( \implies \) \( dt = \sec^{2}x dx \) The integral becomes \( \int \frac { dt }{ (2+t)(3+t) } \) We use partial fractions for \( \frac { 1 }{ (2+t)(3+t) } \): Let \( \frac { 1 }{ (2+t)(3+t) } = \frac { A }{ 2+t } + \frac { B }{ 3+t } \)
\( \implies \) \( 1 = A(3+t) + B(2+t) \) To find A, set \( t = -2 \): \( 1 = A(3-2) + B(2-2) \)
\( \implies \) \( 1 = A(1) \)
\( \implies \) \( A = 1 \) To find B, set \( t = -3 \): \( 1 = A(3-3) + B(2-3) \)
\( \implies \) \( 1 = B(-1) \)
\( \implies \) \( B = -1 \) So, \( \int \frac { dt }{ (2+t)(3+t) } = \int \left( \frac { 1 }{ 2+t } + \frac { -1 }{ 3+t } \right) dt \) \( = \int \frac { 1 }{ 2+t } dt - \int \frac { 1 }{ 3+t } dt \) \( = \log|2+t| - \log|3+t| + C \) \( = \log\left|\frac{2+t}{3+t}\right| + C \) Substitute back \( t = \tan x \): \( = \log\left|\frac{2+\tan x}{3+\tan x}\right| + C \) Recognizing the appropriate substitution helps transform trigonometric integrals into manageable forms.
In simple words: We used a trick by replacing \( \tan x \) with \( t \), which changed the problem into a simpler fraction. Then, we broke this fraction into two easy parts and integrated them. Finally, we put \( \tan x \) back to get the answer.

🎯 Exam Tip: In integrals with \( \sec^2 x \) and functions of \( \tan x \), try the substitution \( t = \tan x \). This is a common pattern that simplifies the integral into a rational function, ready for partial fractions.

Question 21. \( \int \frac { 1 }{ x(x^{5}+1) } dx \)
Answer:To solve this, we can multiply the numerator and denominator by \( x^{4} \): \( \int \frac { x^{4} }{ x^{5}(x^{5}+1) } dx \) Now, let \( t = x^{5} \).
\( \implies \) \( dt = 5x^{4} dx \)
\( \implies \) \( x^{4} dx = \frac{1}{5} dt \) Substitute these into the integral: \( \int \frac { 1 }{ t(t+1) } \frac{1}{5} dt = \frac{1}{5} \int \frac { 1 }{ t(t+1) } dt \) Now, use partial fractions for \( \frac { 1 }{ t(t+1) } \): Let \( \frac { 1 }{ t(t+1) } = \frac { A }{ t } + \frac { B }{ t+1 } \)
\( \implies \) \( 1 = A(t+1) + Bt \) To find A, set \( t = 0 \): \( 1 = A(0+1) + B(0) \)
\( \implies \) \( 1 = A \) To find B, set \( t = -1 \): \( 1 = A(-1+1) + B(-1) \)
\( \implies \) \( 1 = -B \)
\( \implies \) \( B = -1 \) So, \( \frac{1}{5} \int \left( \frac { 1 }{ t } - \frac { 1 }{ t+1 } \right) dt \) \( = \frac{1}{5} \left( \int \frac { 1 }{ t } dt - \int \frac { 1 }{ t+1 } dt \right) \) \( = \frac{1}{5} (\log|t| - \log|t+1|) + C \) \( = \frac{1}{5} \log\left|\frac{t}{t+1}\right| + C \) Substitute back \( t = x^{5} \): \( = \frac{1}{5} \log\left|\frac{x^{5}}{x^{5}+1}\right| + C \) Multiplying the numerator and denominator by a suitable power of x can help set up the substitution correctly.
In simple words: We multiplied the top and bottom of the fraction by \( x^4 \) to make it easier to change \( x^5 \) to \( t \). After replacing, we split the new fraction into two easy parts and integrated them. Finally, we put \( x^5 \) back to finish the problem.

🎯 Exam Tip: For integrals of the form \( \int \frac{1}{x(x^n+1)} dx \), multiply numerator and denominator by \( x^{n-1} \) to facilitate the substitution \( t = x^n \). This is a common trick to make such integrals solvable by partial fractions.

Question 22. \( \int \frac { 1 }{ x(a+bx^{n}) } dx \)
Answer:To solve this integral, we first multiply the numerator and denominator by \( x^{n-1} \): \( \int \frac { x^{n-1} }{ x^{n}(a+bx^{n}) } dx \) Now, let \( t = a+bx^{n} \).
\( \implies \) \( dt = bnx^{n-1} dx \)
\( \implies \) \( x^{n-1} dx = \frac{1}{bn} dt \) Also, from \( t = a+bx^{n} \), we have \( bx^{n} = t-a \), so \( x^{n} = \frac{t-a}{b} \). Substitute these into the integral: \( \int \frac { 1 }{ \frac{t-a}{b} \cdot t } \cdot \frac{1}{bn} dt = \int \frac { b }{ t(t-a)bn } dt = \frac{1}{n} \int \frac { 1 }{ t(t-a) } dt \) Now, use partial fractions for \( \frac { 1 }{ t(t-a) } \): Let \( \frac { 1 }{ t(t-a) } = \frac { A }{ t } + \frac { B }{ t-a } \)
\( \implies \) \( 1 = A(t-a) + Bt \) To find A, set \( t = 0 \): \( 1 = A(0-a) + B(0) \)
\( \implies \) \( 1 = -aA \)
\( \implies \) \( A = -\frac{1}{a} \) To find B, set \( t = a \): \( 1 = A(a-a) + B(a) \)
\( \implies \) \( 1 = Ba \)
\( \implies \) \( B = \frac{1}{a} \) So, \( \frac{1}{n} \int \left( \frac { -1/a }{ t } + \frac { 1/a }{ t-a } \right) dt \) \( = \frac{1}{na} \int \left( \frac { 1 }{ t-a } - \frac { 1 }{ t } \right) dt \) \( = \frac{1}{na} (\log|t-a| - \log|t|) + C \) \( = \frac{1}{na} \log\left|\frac{t-a}{t}\right| + C \) Substitute back \( t = a+bx^{n} \) and \( t-a = bx^{n} \): \( = \frac{1}{na} \log\left|\frac{bx^{n}}{a+bx^{n}}\right| + C \) This systematic approach using substitution and partial fractions is essential for solving such integrals.
In simple words: We made the integral easier by multiplying the top and bottom by \( x^{n-1} \). Then, we replaced \( a+bx^n \) with \( t \). We split the new fraction into two simpler parts and integrated them. Finally, we put \( a+bx^n \) back to get the answer.

🎯 Exam Tip: For integrals of the form \( \int \frac{1}{x(a+bx^n)} dx \), always try to use the substitution \( t = a+bx^n \) after multiplying the numerator and denominator by \( x^{n-1} \). This will simplify the integral into a partial fraction problem.

Question 23. \( \int \frac { 8 }{ (x+2)(x^{2} + 4) } dx \)
Answer:We use partial fractions to simplify the integrand: Let \( \frac { 8 }{ (x+2)(x^{2} + 4) } = \frac { A }{ x+2 } + \frac { Bx+C }{ x^{2} + 4 } \)
\( \implies \) \( 8 = A(x^{2} + 4) + (Bx+C)(x+2) \) To find A, set \( x = -2 \): \( 8 = A((-2)^{2} + 4) + (B(-2)+C)(-2+2) \)
\( \implies \) \( 8 = A(4+4) + 0 \)
\( \implies \) \( 8 = 8A \)
\( \implies \) \( A = 1 \) Now substitute \( A=1 \) back into the equation: \( 8 = 1(x^{2} + 4) + (Bx+C)(x+2) \)
\( \implies \) \( 8 = x^{2} + 4 + Bx^{2} + 2Bx + Cx + 2C \)
\( \implies \) \( 8 = (1+B)x^{2} + (2B+C)x + (4+2C) \) Comparing coefficients: Coefficient of \( x^{2} \): \( 1+B = 0 \implies B = -1 \) Coefficient of \( x \): \( 2B+C = 0 \implies 2(-1)+C = 0 \implies -2+C = 0 \implies C = 2 \) Constant term: \( 4+2C = 8 \implies 4+2(2) = 4+4 = 8 \). This matches. So, \( A = 1, B = -1, C = 2 \). Thus, \( \frac { 8 }{ (x+2)(x^{2} + 4) } = \frac { 1 }{ x+2 } + \frac { -x+2 }{ x^{2} + 4 } = \frac { 1 }{ x+2 } - \frac { x }{ x^{2} + 4 } + \frac { 2 }{ x^{2} + 4 } \) Now, we integrate: \( \int \frac { 8 }{ (x+2)(x^{2} + 4) } dx = \int \frac { 1 }{ x+2 } dx - \int \frac { x }{ x^{2} + 4 } dx + \int \frac { 2 }{ x^{2} + 4 } dx \) For the second term, let \( u = x^{2}+4 \), so \( du = 2x dx \). For the third term, \( \int \frac { 1 }{ x^{2}+a^{2} } dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) \). Here \( a=2 \). \( = \log|x+2| - \frac{1}{2} \int \frac { 2x }{ x^{2} + 4 } dx + 2 \int \frac { 1 }{ x^{2} + 2^{2} } dx \) \( = \log|x+2| - \frac{1}{2} \log|x^{2} + 4| + 2 \cdot \frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right) + C \) \( = \log|x+2| - \frac{1}{2} \log(x^{2} + 4) + \tan^{-1}\left(\frac{x}{2}\right) + C \) This detailed step-by-step approach ensures accurate integration even with complex fractions.
In simple words: We broke the complex fraction into three simpler parts using partial fractions. Then, we integrated each of these simpler parts. The first part gave a logarithm, the second also a logarithm, and the third part gave an inverse tangent.

🎯 Exam Tip: When using partial fractions with irreducible quadratic factors like \( x^{2}+4 \), ensure you use the form \( Bx+C \) in the numerator. Remember to use the standard integral formula for \( \frac{1}{x^2+a^2} \).

Question 24. \( \int \frac { 1-\cos x }{ \cos x(1+\cos x) } dx \)
Answer:We can split the integrand into two terms: \( \frac { 1-\cos x }{ \cos x(1+\cos x) } = \frac { 1 }{ \cos x(1+\cos x) } - \frac { \cos x }{ \cos x(1+\cos x) } \) \( = \frac { 1 }{ \cos x(1+\cos x) } - \frac { 1 }{ 1+\cos x } \) Now, let's work on \( \frac { 1 }{ \cos x(1+\cos x) } \) using partial fractions for \( \frac { 1 }{ u(1+u) } \) where \( u=\cos x \). \( \frac { 1 }{ u(1+u) } = \frac { 1 }{ u } - \frac { 1 }{ 1+u } \) So, \( \frac { 1 }{ \cos x(1+\cos x) } = \frac { 1 }{ \cos x } - \frac { 1 }{ 1+\cos x } \) Substitute this back into the integral: \( \int \left( \frac { 1 }{ \cos x } - \frac { 1 }{ 1+\cos x } - \frac { 1 }{ 1+\cos x } \right) dx \) \( = \int \left( \frac { 1 }{ \cos x } - \frac { 2 }{ 1+\cos x } \right) dx \) \( = \int \sec x dx - 2 \int \frac { 1 }{ 1+\cos x } dx \) We know \( \int \sec x dx = \log|\sec x + \tan x| \). For \( \int \frac { 1 }{ 1+\cos x } dx \), we use the half-angle identity \( 1+\cos x = 2\cos^{2}\left(\frac{x}{2}\right) \): \( \int \frac { 1 }{ 2\cos^{2}\left(\frac{x}{2}\right) } dx = \frac{1}{2} \int \sec^{2}\left(\frac{x}{2}\right) dx \) Let \( y = \frac{x}{2} \), so \( dy = \frac{1}{2} dx \implies dx = 2dy \). \( \frac{1}{2} \int \sec^{2}(y) (2dy) = \int \sec^{2}(y) dy = \tan y + K = \tan\left(\frac{x}{2}\right) + K \) Combining all parts: \( \int \frac { 1-\cos x }{ \cos x(1+\cos x) } dx = \log|\sec x + \tan x| - 2 \tan\left(\frac{x}{2}\right) + C \) Splitting complex fractions and using trigonometric identities are powerful techniques in integration.
In simple words: We first broke the fraction into two simpler parts. Then, we used another trick to split one of these parts even further. After that, we integrated each piece separately using known formulas and trigonometric identities to get the final answer.

🎯 Exam Tip: For integrals involving \( \cos x \) or \( \sin x \) in the denominator, look for ways to use half-angle formulas (like \( 1+\cos x = 2\cos^2(x/2) \)) or substitution to simplify the expression into standard integral forms.