RBSE Solutions Class 12 Maths Chapter 9 Integration Exercise 9.4

Ex 9.4

Integrate The Following Functions W.R.T. X.

Question 1. \( \frac { 1 }{ 16 - 9x^2 } \)
Answer: We need to integrate \( \frac { 1 }{ 16 - 9x^2 } \) with respect to x. First, rewrite the denominator to match a standard integration formula.
\( \int \frac { 1 }{ 16 - 9x^2 } dx \)
\( = \int \frac { 1 }{ 4^2 - (3x)^2 } dx \)
Let \( u = 3x \). Then \( du = 3 dx \), so \( dx = \frac{1}{3} du \).
\( = \int \frac { 1 }{ 4^2 - u^2 } \frac{1}{3} du \)
\( = \frac{1}{3} \int \frac { 1 }{ 4^2 - u^2 } du \)
Using the formula \( \int \frac { 1 }{ a^2 - x^2 } dx = \frac { 1 }{ 2a } \log \left| \frac { a+x }{ a-x } \right| + C \), here \( a=4 \) and \( x=u \):
\( = \frac{1}{3} \left[ \frac { 1 }{ 2(4) } \log \left| \frac { 4+u }{ 4-u } \right| \right] + C \)
Substitute \( u=3x \) back:
\( = \frac{1}{3} \left[ \frac { 1 }{ 8 } \log \left| \frac { 4+3x }{ 4-3x } \right| \right] + C \)
\( = \frac { 1 }{ 24 } \log \left| \frac { 4+3x }{ 4-3x } \right| + C \)
In simple words: To solve this, we first change the bottom part of the fraction to fit a known math rule. Then we replace a part of the expression with 'u' to make it simpler, and use a special formula for fractions with squares to find the final integral.

🎯 Exam Tip: Remember to adjust for the derivative of the inner function (like 3x in this case) by multiplying or dividing by its derivative when performing substitution.

Question 2. \( \frac { 1 }{ x^2 - 36 } \)
Answer: We need to integrate the given function. First, we rewrite the denominator as a difference of squares.
\( \int \frac { 1 }{ x^2 - 36 } dx \)
\( = \int \frac { 1 }{ x^2 - 6^2 } dx \)
Using the standard integration formula \( \int \frac { 1 }{ x^2 - a^2 } dx = \frac { 1 }{ 2a } \log \left| \frac { x-a }{ x+a } \right| + C \), where \( a=6 \):
\( = \frac { 1 }{ 2(6) } \log \left| \frac { x-6 }{ x+6 } \right| + C \)
\( = \frac { 1 }{ 12 } \log \left| \frac { x-6 }{ x+6 } \right| + C \)
In simple words: This problem asks us to find the integral of a fraction. We change the bottom part to look like a square number minus another square number. Then we use a specific formula to solve it quickly.

🎯 Exam Tip: Always recognize the basic integration formulas for functions involving \( a^2 \pm x^2 \) or \( x^2 \pm a^2 \) in the denominator, as they are frequently used.

Question 3. \( \frac { 3x }{ (x+1)(x-2) } \)
Answer: We need to integrate this rational function using partial fraction decomposition. First, we express the fraction as a sum of simpler fractions.
Let \( \frac { 3x }{ (x+1)(x-2) } = \frac { A }{ x+1 } + \frac { B }{ x-2 } \)
Multiply both sides by \( (x+1)(x-2) \):
\( 3x = A(x-2) + B(x+1) \)
To find A, set \( x = -1 \):
\( 3(-1) = A(-1-2) + B(-1+1) \)
\( -3 = -3A \implies A=1 \)
To find B, set \( x = 2 \):
\( 3(2) = A(2-2) + B(2+1) \)
\( 6 = 3B \implies B=2 \)
So, the original expression can be written as:
\( \frac { 3x }{ (x+1)(x-2) } = \frac { 1 }{ x+1 } + \frac { 2 }{ x-2 } \)
Now, we integrate this sum:
\( \int \frac { 3x }{ (x+1)(x-2) } dx = \int \left( \frac { 1 }{ x+1 } + \frac { 2 }{ x-2 } \right) dx \)
\( = \int \frac { 1 }{ x+1 } dx + 2 \int \frac { 1 }{ x-2 } dx \)
\( = \log |x+1| + 2 \log |x-2| + C \)
In simple words: We break down the complex fraction into simpler parts. Then, we find the values for these new parts and integrate each one separately. This makes a difficult fraction easy to integrate.

🎯 Exam Tip: When using partial fractions, always make sure the degree of the numerator is less than the degree of the denominator. If not, perform polynomial long division first.

Question 4. \( \frac { 3x-2 }{ (x+1)^2(x+3) } \)
Answer: We will use partial fraction decomposition to integrate this expression. Since there is a repeated factor \( (x+1)^2 \), we set up the partial fractions as follows:
Let \( \frac { 3x-2 }{ (x+1)^2(x+3) } = \frac { A }{ x+1 } + \frac { B }{ (x+1)^2 } + \frac { C }{ x+3 } \)
Multiply both sides by \( (x+1)^2(x+3) \):
\( 3x-2 = A(x+1)(x+3) + B(x+3) + C(x+1)^2 \)
\( 3x-2 = A(x^2+4x+3) + B(x+3) + C(x^2+2x+1) \)
\( 3x-2 = Ax^2+4Ax+3A + Bx+3B + Cx^2+2Cx+C \)
Group terms by powers of x:
\( 3x-2 = (A+C)x^2 + (4A+B+2C)x + (3A+3B+C) \)
Comparing the coefficients of like powers of x on both sides:
For \( x^2 \): \( A+C=0 \) ...(i)
For \( x \): \( 4A+B+2C=3 \) ...(ii)
For constant: \( 3A+3B+C=-2 \) ...(iii)
From (i), \( C=-A \). Substitute this into (ii) and (iii):
\( 4A+B+2(-A)=3 \implies 2A+B=3 \) ...(iv)
\( 3A+3B+(-A)=-2 \implies 2A+3B=-2 \) ...(v)
Subtract equation (iv) from (v):
\( (2A+3B)-(2A+B) = -2-3 \)
\( 2B = -5 \implies B = -\frac{5}{2} \)
Substitute \( B = -\frac{5}{2} \) into (iv):
\( 2A - \frac{5}{2} = 3 \)
\( 2A = 3 + \frac{5}{2} = \frac{6+5}{2} = \frac{11}{2} \)
\( A = \frac{11}{4} \)
Since \( C=-A \), \( C = -\frac{11}{4} \)
Now, substitute A, B, C back into the partial fraction form and integrate:
\( \int \left( \frac { \frac{11}{4} }{ x+1 } + \frac { -\frac{5}{2} }{ (x+1)^2 } + \frac { -\frac{11}{4} }{ x+3 } \right) dx \)
\( = \frac{11}{4} \int \frac { 1 }{ x+1 } dx - \frac{5}{2} \int (x+1)^{-2} dx - \frac{11}{4} \int \frac { 1 }{ x+3 } dx \)
\( = \frac{11}{4} \log |x+1| - \frac{5}{2} \left( \frac{(x+1)^{-1}}{-1} \right) - \frac{11}{4} \log |x+3| + K \)
\( = \frac{11}{4} \log |x+1| + \frac{5}{2(x+1)} - \frac{11}{4} \log |x+3| + K \)
This can also be written as \( \frac{11}{4} \log \left| \frac{x+1}{x+3} \right| + \frac{5}{2(x+1)} + K \). This method simplifies integration of complex fractions.
In simple words: We break the fraction into three simpler ones because one part of the bottom is squared. We find the unknown numbers for these new fractions by comparing the top parts. Then we integrate each simple fraction to get the final answer.

🎯 Exam Tip: For repeated linear factors like \( (ax+b)^n \), remember to include terms for all powers from 1 to n in your partial fraction setup, i.e., \( \frac{A_1}{ax+b} + \frac{A_2}{(ax+b)^2} + \dots + \frac{A_n}{(ax+b)^n} \).

Question 5. \( \frac { x^2 }{ (x+1)(x-2)(x-3) } \)
Answer: We will solve this integral using the partial fraction method. We break down the complex fraction into simpler ones with linear denominators.
Let \( \frac { x^2 }{ (x+1)(x-2)(x-3) } = \frac { A }{ x+1 } + \frac { B }{ x-2 } + \frac { C }{ x-3 } \)
Multiply both sides by \( (x+1)(x-2)(x-3) \):
\( x^2 = A(x-2)(x-3) + B(x+1)(x-3) + C(x+1)(x-2) \)
To find A, set \( x = -1 \):
\( (-1)^2 = A(-1-2)(-1-3) \)
\( 1 = A(-3)(-4) \implies 1 = 12A \implies A = \frac{1}{12} \)
To find B, set \( x = 2 \):
\( (2)^2 = B(2+1)(2-3) \)
\( 4 = B(3)(-1) \implies 4 = -3B \implies B = -\frac{4}{3} \)
To find C, set \( x = 3 \):
\( (3)^2 = C(3+1)(3-2) \)
\( 9 = C(4)(1) \implies 9 = 4C \implies C = \frac{9}{4} \)
Now, substitute A, B, C back into the partial fraction form and integrate:
\( \int \left( \frac { \frac{1}{12} }{ x+1 } + \frac { -\frac{4}{3} }{ x-2 } + \frac { \frac{9}{4} }{ x-3 } \right) dx \)
\( = \frac{1}{12} \int \frac { 1 }{ x+1 } dx - \frac{4}{3} \int \frac { 1 }{ x-2 } dx + \frac{9}{4} \int \frac { 1 }{ x-3 } dx \)
\( = \frac{1}{12} \log |x+1| - \frac{4}{3} \log |x-2| + \frac{9}{4} \log |x-3| + K \)
This process is effective for fractions where the denominator can be factored into distinct linear factors.
In simple words: We split the fraction into three simpler parts. By picking special 'x' values, we find the numbers for each part. Then, we integrate each simple part, which usually means taking the logarithm of the denominator.

🎯 Exam Tip: When using the cover-up method (setting x to the roots of the denominator) to find coefficients, ensure you apply it carefully to each factor to avoid errors.

Question 6. \( \frac { x^2 }{ x^4 - x^2 - 12 } \)
Answer: To solve this integral, we first simplify the expression using substitution and then apply partial fraction decomposition.
Let \( y = x^2 \). Then the expression becomes \( \frac{y}{y^2-y-12} \).
Factor the denominator: \( y^2-y-12 = (y-4)(y+3) \).
So, we have \( \frac{y}{(y-4)(y+3)} \).
Now, we decompose this into partial fractions:
Let \( \frac { y }{ (y-4)(y+3) } = \frac { A }{ y-4 } + \frac { B }{ y+3 } \)
Multiply both sides by \( (y-4)(y+3) \):
\( y = A(y+3) + B(y-4) \)
To find A, set \( y=4 \):
\( 4 = A(4+3) + B(4-4) \)
\( 4 = 7A \implies A = \frac{4}{7} \)
To find B, set \( y=-3 \):
\( -3 = A(-3+3) + B(-3-4) \)
\( -3 = -7B \implies B = \frac{3}{7} \)
So, \( \frac { y }{ (y-4)(y+3) } = \frac { \frac{4}{7} }{ y-4 } + \frac { \frac{3}{7} }{ y+3 } \)
Substitute back \( y = x^2 \):
\( \frac { x^2 }{ x^4-x^2-12 } = \frac { \frac{4}{7} }{ x^2-4 } + \frac { \frac{3}{7} }{ x^2+3 } \)
Now, we integrate the expression:
\( \int \frac { x^2 }{ x^4-x^2-12 } dx = \int \left( \frac { \frac{4}{7} }{ x^2-4 } + \frac { \frac{3}{7} }{ x^2+3 } \right) dx \)
\( = \frac{4}{7} \int \frac { 1 }{ x^2-2^2 } dx + \frac{3}{7} \int \frac { 1 }{ x^2+(\sqrt{3})^2 } dx \)
Using the standard formulas: \( \int \frac { 1 }{ x^2-a^2 } dx = \frac { 1 }{ 2a } \log \left| \frac { x-a }{ x+a } \right| \) and \( \int \frac { 1 }{ x^2+a^2 } dx = \frac { 1 }{ a } \tan^{-1} \left( \frac{x}{a} \right) \)
\( = \frac{4}{7} \left( \frac { 1 }{ 2(2) } \log \left| \frac { x-2 }{ x+2 } \right| \right) + \frac{3}{7} \left( \frac { 1 }{ \sqrt{3} } \tan^{-1} \left( \frac{x}{\sqrt{3}} \right) \right) + K \)
\( = \frac{4}{28} \log \left| \frac { x-2 }{ x+2 } \right| + \frac{3}{7\sqrt{3}} \tan^{-1} \left( \frac{x}{\sqrt{3}} \right) + K \)
\( = \frac{1}{7} \log \left| \frac { x-2 }{ x+2 } \right| + \frac{\sqrt{3}}{7} \tan^{-1} \left( \frac{x}{\sqrt{3}} \right) + K \)
This method helps break down complex polynomial fractions into simpler, integrable forms.
In simple words: First, we change \( x^2 \) to 'y' to make the fraction simpler. Then, we split this new fraction into two simpler parts. After finding the values for these parts, we put \( x^2 \) back and integrate each part using known rules for fractions with squares.

🎯 Exam Tip: For integrals involving \( x^{2n} \), consider a substitution like \( y = x^n \) to simplify the expression, especially when the denominator can be factored using this substitution.

Question 7. \( \frac { 1 }{ x^3 - x^2 - x + 1 } \)
Answer: We need to integrate this function. First, we factor the denominator to apply partial fractions.
Factor the denominator:
\( x^3-x^2-x+1 = x^2(x-1) - 1(x-1) \)
\( = (x^2-1)(x-1) \)
\( = (x-1)(x+1)(x-1) \)
\( = (x-1)^2(x+1) \)
So the integral becomes: \( \int \frac { 1 }{ (x-1)^2(x+1) } dx \)
Now, we decompose this into partial fractions. Since \( (x-1)^2 \) is a repeated linear factor, we set up the fractions as:
Let \( \frac { 1 }{ (x-1)^2(x+1) } = \frac { A }{ x-1 } + \frac { B }{ (x-1)^2 } + \frac { C }{ x+1 } \)
Multiply both sides by \( (x-1)^2(x+1) \):
\( 1 = A(x-1)(x+1) + B(x+1) + C(x-1)^2 \)
To find B, set \( x=1 \):
\( 1 = A(0) + B(1+1) + C(0) \)
\( 1 = 2B \implies B = \frac{1}{2} \)
To find C, set \( x=-1 \):
\( 1 = A(0) + B(0) + C(-1-1)^2 \)
\( 1 = C(-2)^2 \implies 1 = 4C \implies C = \frac{1}{4} \)
To find A, compare coefficients or set \( x=0 \):
Using \( x=0 \):
\( 1 = A(0-1)(0+1) + B(0+1) + C(0-1)^2 \)
\( 1 = -A + B + C \)
\( 1 = -A + \frac{1}{2} + \frac{1}{4} \)
\( 1 = -A + \frac{3}{4} \)
\( A = \frac{3}{4} - 1 = -\frac{1}{4} \)
Now, substitute A, B, C back into the partial fraction form and integrate:
\( \int \left( \frac { -\frac{1}{4} }{ x-1 } + \frac { \frac{1}{2} }{ (x-1)^2 } + \frac { \frac{1}{4} }{ x+1 } \right) dx \)
\( = -\frac{1}{4} \int \frac { 1 }{ x-1 } dx + \frac{1}{2} \int (x-1)^{-2} dx + \frac{1}{4} \int \frac { 1 }{ x+1 } dx \)
\( = -\frac{1}{4} \log |x-1| + \frac{1}{2} \left( \frac{(x-1)^{-1}}{-1} \right) + \frac{1}{4} \log |x+1| + K \)
\( = -\frac{1}{4} \log |x-1| - \frac{1}{2(x-1)} + \frac{1}{4} \log |x+1| + K \)
\( = \frac{1}{4} \log \left| \frac{x+1}{x-1} \right| - \frac{1}{2(x-1)} + K \)
Factoring the denominator is a key step in simplifying this type of rational function integration.
In simple words: First, we break down the bottom part of the fraction into its smaller multiplication parts. Then, we use a special method called partial fractions because one part is squared. We find the numbers for each simple fraction and then add up their integrals.

🎯 Exam Tip: Always look for ways to factorize polynomial denominators completely before applying partial fraction decomposition. Synthetic division or grouping can often help.

Question 8. \( \int \frac{x^2}{(x+1)(x-2)} dx \)
Answer: We need to integrate this rational function. Since the degree of the numerator (2) is equal to the degree of the denominator (2), we must perform polynomial long division first.
First, expand the denominator: \( (x+1)(x-2) = x^2 - 2x + x - 2 = x^2 - x - 2 \).
Now, perform the division:
\( \frac{x^2}{x^2-x-2} = 1 + \frac{x+2}{x^2-x-2} \)
So, the integral becomes:
\( \int \left( 1 + \frac{x+2}{(x+1)(x-2)} \right) dx = \int 1 dx + \int \frac{x+2}{(x+1)(x-2)} dx \)
Now, we decompose the fractional part into partial fractions:
Let \( \frac { x+2 }{ (x+1)(x-2) } = \frac { A }{ x+1 } + \frac { B }{ x-2 } \)
Multiply both sides by \( (x+1)(x-2) \):
\( x+2 = A(x-2) + B(x+1) \)
To find A, set \( x=-1 \):
\( -1+2 = A(-1-2) + B(-1+1) \)
\( 1 = -3A \implies A = -\frac{1}{3} \)
To find B, set \( x=2 \):
\( 2+2 = A(2-2) + B(2+1) \)
\( 4 = 3B \implies B = \frac{4}{3} \)
So, the fractional part is:
\( \frac { x+2 }{ (x+1)(x-2) } = \frac { -\frac{1}{3} }{ x+1 } + \frac { \frac{4}{3} }{ x-2 } \)
Now, integrate this sum along with the '1' from the division:
\( \int \left( 1 - \frac{1}{3(x+1)} + \frac{4}{3(x-2)} \right) dx \)
\( = \int 1 dx - \frac{1}{3} \int \frac { 1 }{ x+1 } dx + \frac{4}{3} \int \frac { 1 }{ x-2 } dx \)
\( = x - \frac{1}{3} \log |x+1| + \frac{4}{3} \log |x-2| + K \)
This method helps integrate rational functions where the numerator's degree is not less than the denominator's.
In simple words: First, we divide the top by the bottom because they have the same power. This gives us a whole number and a new fraction. Then, we break this new fraction into two simpler parts. Finally, we integrate the whole number and each simple part separately.

🎯 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. \( \frac { x^2 }{ (x^2+a^2)(x^2+b^2) } \)"
Answer: We solve this integral by first using a substitution and then applying partial fraction decomposition.
Let \( y = x^2 \). Then the expression becomes \( \frac{y}{(y+a^2)(y+b^2)} \).
Now, decompose this into partial fractions:
Let \( \frac { y }{ (y+a^2)(y+b^2) } = \frac { A }{ y+a^2 } + \frac { B }{ y+b^2 } \)
Multiply both sides by \( (y+a^2)(y+b^2) \):
\( y = A(y+b^2) + B(y+a^2) \)
To find A, set \( y=-a^2 \):
\( -a^2 = A(-a^2+b^2) + B(-a^2+a^2) \)
\( -a^2 = A(b^2-a^2) \implies A = \frac{-a^2}{b^2-a^2} = \frac{a^2}{a^2-b^2} \)
To find B, set \( y=-b^2 \):
\( -b^2 = A(-b^2+b^2) + B(-b^2+a^2) \)
\( -b^2 = B(a^2-b^2) \implies B = \frac{-b^2}{a^2-b^2} \)
So, the expression is:
\( \frac { y }{ (y+a^2)(y+b^2) } = \frac { \frac{a^2}{a^2-b^2} }{ y+a^2 } + \frac { \frac{-b^2}{a^2-b^2} }{ y+b^2 } \)
Substitute back \( y = x^2 \):
\( \frac { x^2 }{ (x^2+a^2)(x^2+b^2) } = \frac{1}{a^2-b^2} \left( \frac { a^2 }{ x^2+a^2 } - \frac { b^2 }{ x^2+b^2 } \right) \)
Now, we integrate the expression:
\( \int \frac { x^2 }{ (x^2+a^2)(x^2+b^2) } dx = \frac{1}{a^2-b^2} \left( a^2 \int \frac { 1 }{ x^2+a^2 } dx - b^2 \int \frac { 1 }{ x^2+b^2 } dx \right) \)
Using the standard integration formula \( \int \frac { 1 }{ x^2+c^2 } dx = \frac { 1 }{ c } \tan^{-1} \left( \frac{x}{c} \right) \):
\( = \frac{1}{a^2-b^2} \left( a^2 \left( \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) \right) - b^2 \left( \frac{1}{b} \tan^{-1} \left( \frac{x}{b} \right) \right) \right) + C \)
\( = \frac{1}{a^2-b^2} \left( a \tan^{-1} \left( \frac{x}{a} \right) - b \tan^{-1} \left( \frac{x}{b} \right) \right) + C \)
This type of problem requires careful handling of algebraic constants in the partial fraction setup.
In simple words: We first replace \( x^2 \) with a temporary letter 'y'. Then we split the fraction into two simpler parts involving 'y'. We find the values for these parts. Finally, we put \( x^2 \) back and integrate each part using a rule that involves the inverse tangent function.

🎯 Exam Tip: When dealing with factors like \( (x^2+a^2) \), the corresponding partial fraction term should be of the form \( \frac{Ax+B}{x^2+a^2} \). However, if the numerator is also \( x^2 \), a substitution like \( y=x^2 \) can sometimes simplify the problem to linear factors in 'y', making it easier.

Question 10. \( \frac { x+1 }{ x^3 + x^2 - 6x } \)
Answer: To integrate this rational function, we first factorize the denominator completely and then use partial fraction decomposition.
Factor the denominator:
\( x^3+x^2-6x = x(x^2+x-6) \)
\( = x(x+3)(x-2) \)
So the integral becomes: \( \int \frac{x+1}{x(x+3)(x-2)} dx \)
Now, we decompose this into partial fractions:
Let \( \frac { x+1 }{ x(x+3)(x-2) } = \frac { A }{ x } + \frac { B }{ x+3 } + \frac { C }{ x-2 } \)
Multiply both sides by \( x(x+3)(x-2) \):
\( x+1 = A(x+3)(x-2) + Bx(x-2) + Cx(x+3) \)
To find A, set \( x=0 \):
\( 0+1 = A(0+3)(0-2) + B(0) + C(0) \)
\( 1 = A(3)(-2) \implies 1 = -6A \implies A = -\frac{1}{6} \)
To find B, set \( x=-3 \):
\( -3+1 = A(0) + B(-3)(-3-2) + C(0) \)
\( -2 = B(-3)(-5) \implies -2 = 15B \implies B = -\frac{2}{15} \)
To find C, set \( x=2 \):
\( 2+1 = A(0) + B(0) + C(2)(2+3) \)
\( 3 = C(2)(5) \implies 3 = 10C \implies C = \frac{3}{10} \)
Now, substitute A, B, C back into the partial fraction form and integrate:
\( \int \left( \frac { -\frac{1}{6} }{ x } + \frac { -\frac{2}{15} }{ x+3 } + \frac { \frac{3}{10} }{ x-2 } \right) dx \)
\( = -\frac{1}{6} \int \frac { 1 }{ x } dx - \frac{2}{15} \int \frac { 1 }{ x+3 } dx + \frac{3}{10} \int \frac { 1 }{ x-2 } dx \)
\( = -\frac{1}{6} \log |x| - \frac{2}{15} \log |x+3| + \frac{3}{10} \log |x-2| + K \)
This method efficiently breaks down complex rational functions into simpler, integrable logarithmic terms.
In simple words: First, we find the individual parts that make up the bottom of the fraction by factoring. Then, we use the partial fractions method to split the big fraction into smaller, easier pieces. After finding the numbers for these pieces, we integrate each simple fraction, which gives us a sum of logarithm terms.

🎯 Exam Tip: Remember to factor out common terms like 'x' from the denominator first, which often reveals simpler factors for partial fraction decomposition.

Question 11. \( \int \frac{x^2-8x+4}{x^3-4x} dx \)
Answer: We need to evaluate this integral. First, we factorize the denominator completely and then use partial fraction decomposition.
Factor the denominator:
\( x^3-4x = x(x^2-4) \)
\( = x(x-2)(x+2) \)
So the integral becomes: \( \int \frac{x^2-8x+4}{x(x-2)(x+2)} dx \)
Now, we decompose this into partial fractions:
Let \( \frac { x^2-8x+4 }{ x(x-2)(x+2) } = \frac { A }{ x } + \frac { B }{ x-2 } + \frac { C }{ x+2 } \)
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 A, set \( x=0 \):
\( (0)^2-8(0)+4 = A(0-2)(0+2) \)
\( 4 = A(-2)(2) \implies 4 = -4A \implies A = -1 \)
To find B, set \( x=2 \):
\( (2)^2-8(2)+4 = B(2)(2+2) \)
\( 4-16+4 = B(2)(4) \implies -8 = 8B \implies B = -1 \)
To find C, set \( x=-2 \):
\( (-2)^2-8(-2)+4 = C(-2)(-2-2) \)
\( 4+16+4 = C(-2)(-4) \implies 24 = 8C \implies C = 3 \)
Now, substitute A, B, C back into the partial fraction form and integrate:
\( \int \left( \frac { -1 }{ x } + \frac { -1 }{ x-2 } + \frac { 3 }{ x+2 } \right) dx \)
\( = -1 \int \frac { 1 }{ x } dx - 1 \int \frac { 1 }{ x-2 } dx + 3 \int \frac { 1 }{ x+2 } dx \)
\( = -\log |x| - \log |x-2| + 3 \log |x+2| + K \)
This method effectively simplifies complex rational integrals into a sum of basic logarithmic integrals.
In simple words: First, we find all the simple parts of the bottom of the fraction by factoring it. Then, we use partial fractions to break the main fraction into three simpler ones. After finding the numbers for these parts, we integrate each simple fraction separately to get the final answer.

🎯 Exam Tip: Always completely factorize the denominator before setting up partial fractions, identifying all distinct and repeated linear or irreducible quadratic factors.

Question 12. \( \frac { 1 }{ (x-1)^2(x+2) } \)
Answer: We need to integrate this rational function using partial fraction decomposition, as it has a repeated linear factor in the denominator.
We set up the partial fractions as follows:
Let \( \frac { 1 }{ (x-1)^2(x+2) } = \frac { A }{ x-1 } + \frac { B }{ (x-1)^2 } + \frac { C }{ x+2 } \)
Multiply both sides by \( (x-1)^2(x+2) \):
\( 1 = A(x-1)(x+2) + B(x+2) + C(x-1)^2 \)
To find B, set \( x=1 \):
\( 1 = A(0) + B(1+2) + C(0) \)
\( 1 = 3B \implies B = \frac{1}{3} \)
To find C, set \( x=-2 \):
\( 1 = A(0) + B(0) + C(-2-1)^2 \)
\( 1 = C(-3)^2 \implies 1 = 9C \implies C = \frac{1}{9} \)
To find A, compare coefficients of \( x^2 \):
Expand the right side of \( 1 = A(x-1)(x+2) + B(x+2) + C(x-1)^2 \):
\( 1 = A(x^2+x-2) + B(x+2) + C(x^2-2x+1) \)
\( 1 = Ax^2+Ax-2A + Bx+2B + Cx^2-2Cx+C \)
\( 1 = (A+C)x^2 + (A+B-2C)x + (-2A+2B+C) \)
Comparing coefficients of \( x^2 \): \( A+C=0 \implies A = -C \).
Since \( C=\frac{1}{9} \), \( A = -\frac{1}{9} \).
Now, substitute A, B, C back into the partial fraction form and integrate:
\( \int \left( \frac { -\frac{1}{9} }{ x-1 } + \frac { \frac{1}{3} }{ (x-1)^2 } + \frac { \frac{1}{9} }{ x+2 } \right) dx \)
\( = -\frac{1}{9} \int \frac { 1 }{ x-1 } dx + \frac{1}{3} \int (x-1)^{-2} dx + \frac{1}{9} \int \frac { 1 }{ x+2 } dx \)
\( = -\frac{1}{9} \log |x-1| + \frac{1}{3} \left( -\frac{1}{x-1} \right) + \frac{1}{9} \log |x+2| + K \)
\( = -\frac{1}{9} \log |x-1| - \frac{1}{3(x-1)} + \frac{1}{9} \log |x+2| + K \)
This can also be written as \( \frac{1}{9} \log \left| \frac{x+2}{x-1} \right| - \frac{1}{3(x-1)} + K \). This technique is crucial for integrals with repeated factors.
In simple words: We split the fraction into three simpler parts because one factor at the bottom is squared. We find the numbers for each part using special 'x' values or by matching the coefficients. Then, we integrate each simple fraction, which involves logarithms and a simple power rule.

🎯 Exam Tip: When setting up partial fractions for a repeated factor like \( (ax+b)^n \), remember to include terms for each power from 1 up to n. For instance, for \( (x-1)^2 \), you need both \( \frac{A}{x-1} \) and \( \frac{B}{(x-1)^2} \).

Question 13. \( \frac { 1-3x }{ 1+x+x^2+x^3 } \)
Answer: To evaluate this integral, we first factorize the denominator and then apply partial fraction decomposition, which involves an irreducible quadratic factor.
Factor the denominator:
\( 1+x+x^2+x^3 = 1(1+x) + x^2(1+x) \)
\( = (1+x)(1+x^2) \)
So the integral becomes: \( \int \frac{1-3x}{(1+x)(1+x^2)} dx \)
Now, we decompose this into partial fractions:
Let \( \frac { 1-3x }{ (1+x)(1+x^2) } = \frac { A }{ 1+x } + \frac { Bx+C }{ 1+x^2 } \)
Multiply both sides by \( (1+x)(1+x^2) \):
\( 1-3x = A(1+x^2) + (Bx+C)(1+x) \)
To find A, set \( x=-1 \):
\( 1-3(-1) = A(1+(-1)^2) + (B(-1)+C)(1-1) \)
\( 1+3 = A(2) + 0 \implies 4 = 2A \implies A = 2 \)
Now, expand the right side and compare coefficients to find B and C:
\( 1-3x = A + Ax^2 + Bx + Bx^2 + C + Cx \)
\( 1-3x = (A+B)x^2 + (B+C)x + (A+C) \)
Comparing coefficients of \( x^2 \): \( A+B=0 \)
Since \( A=2 \), \( 2+B=0 \implies B=-2 \).
Comparing coefficients of \( x \): \( B+C=-3 \)
Since \( B=-2 \), \( -2+C=-3 \implies C=-1 \).
(We can verify with the constant term: \( A+C=1 \implies 2+(-1)=1 \), which is correct.)
So, the partial fraction decomposition is:
\( \frac { 1-3x }{ (1+x)(1+x^2) } = \frac { 2 }{ 1+x } + \frac { -2x-1 }{ 1+x^2 } = \frac { 2 }{ 1+x } - \frac { 2x+1 }{ 1+x^2 } \)
Now, integrate the expression:
\( \int \left( \frac { 2 }{ 1+x } - \frac { 2x+1 }{ 1+x^2 } \right) dx \)
\( = 2 \int \frac { 1 }{ 1+x } dx - \int \frac { 2x }{ 1+x^2 } dx - \int \frac { 1 }{ 1+x^2 } dx \)
\( = 2 \log |1+x| - \log |1+x^2| - \tan^{-1}(x) + K \)
This method is essential for integrals involving quadratic factors that cannot be further broken down into linear factors.
In simple words: We first find the factors of the bottom part of the fraction, one of which will be a squared term that cannot be factored more. Then, we break the fraction into simpler parts, one for the simple factor and one for the squared factor. We find the numbers for these parts and then integrate each part using logarithms and inverse tangent functions.

🎯 Exam Tip: For irreducible quadratic factors \( (ax^2+bx+c) \) in the denominator, the corresponding partial fraction term should be of the form \( \frac{Ax+B}{ax^2+bx+c} \). Remember to split the numerator \( (Ax+B) \) for easier integration.

Question 14. \( \frac { 1+x^2 }{ x^5-x } \)
Answer: We need to integrate this rational function. First, we factorize the denominator completely and simplify the expression.
Factor the denominator:
\( x^5-x = x(x^4-1) \)
\( = x(x^2-1)(x^2+1) \)
\( = x(x-1)(x+1)(x^2+1) \)
Now, substitute this back into the original expression:
\( \frac { 1+x^2 }{ x(x-1)(x+1)(x^2+1) } \)
We can cancel out the common factor \( (1+x^2) \) from the numerator and denominator:
\( = \frac { 1 }{ x(x-1)(x+1) } \)
Expand the denominator: \( x(x-1)(x+1) = x(x^2-1) = x^3-x \).
So, the integral simplifies to: \( \int \frac { 1 }{ x^3-x } dx \).
Now, we decompose this into partial fractions:
Let \( \frac { 1 }{ x(x-1)(x+1) } = \frac { A }{ x } + \frac { B }{ x-1 } + \frac { C }{ x+1 } \)
Multiply both sides by \( x(x-1)(x+1) \):
\( 1 = A(x-1)(x+1) + Bx(x+1) + Cx(x-1) \)
To find A, set \( x=0 \):
\( 1 = A(0-1)(0+1) + B(0) + C(0) \)
\( 1 = A(-1)(1) \implies 1 = -A \implies A = -1 \)
To find B, set \( x=1 \):
\( 1 = A(0) + B(1)(1+1) + C(0) \)
\( 1 = B(1)(2) \implies 1 = 2B \implies B = \frac{1}{2} \)
To find C, set \( x=-1 \):
\( 1 = A(0) + B(0) + C(-1)(-1-1) \)
\( 1 = C(-1)(-2) \implies 1 = 2C \implies C = \frac{1}{2} \)
Now, substitute A, B, C back into the partial fraction form and integrate:
\( \int \left( \frac { -1 }{ x } + \frac { \frac{1}{2} }{ x-1 } + \frac { \frac{1}{2} }{ x+1 } \right) dx \)
\( = -1 \int \frac { 1 }{ x } dx + \frac{1}{2} \int \frac { 1 }{ x-1 } dx + \frac{1}{2} \int \frac { 1 }{ x+1 } dx \)
\( = -\log |x| + \frac{1}{2} \log |x-1| + \frac{1}{2} \log |x+1| + K \)
This can also be written as \( -\log |x| + \frac{1}{2} \log |(x-1)(x+1)| + K = \log \left| \frac{\sqrt{x^2-1}}{x} \right| + K \). Simplifying the expression before integration can make the problem much easier.
In simple words: First, we break down the bottom of the fraction into all its smallest multiplication parts. We then notice that some parts cancel out, making the fraction much simpler. After that, we break the new simple fraction into even smaller parts. Finally, we integrate these small parts, which mostly results in logarithm terms.

🎯 Exam Tip: Always look for common factors in the numerator and denominator that can be cancelled out before applying partial fraction decomposition. This simplification can significantly reduce the complexity of the problem.

Question 15. Integrate the following function: \( \frac{x^2+5x+3}{x^2+3x+2} \)
Answer: To integrate the given rational function, we first rewrite the fraction because the degree of the numerator is equal to the degree of the denominator. \[ \frac{x^2+5x+3}{x^2+3x+2} = \frac{(x^2+3x+2) + (2x+1)}{x^2+3x+2} = 1 + \frac{2x+1}{x^2+3x+2} \] Now, we factor the denominator of the second term: \( x^2+3x+2 = (x+1)(x+2) \). Next, we use partial fractions for \( \frac{2x+1}{(x+1)(x+2)} \): \( \frac{2x+1}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2} \) Multiply both sides by \( (x+1)(x+2) \): \( 2x+1 = A(x+2) + B(x+1) \) To find A, set \( x=-1 \): \( 2(-1)+1 = A(-1+2) \implies -1 = A \). To find B, set \( x=-2 \): \( 2(-2)+1 = B(-2+1) \implies -3 = -B \implies B=3 \). So the integral becomes: \( \int \left(1 + \frac{-1}{x+1} + \frac{3}{x+2}\right) dx \) \( = \int 1 \,dx - \int \frac{1}{x+1} \,dx + \int \frac{3}{x+2} \,dx \) \( = x - \log|x+1| + 3 \log|x+2| + C \) Here, the constant of integration \( C \) is added at the end.
In simple words: First, split the fraction into a whole number part and a simpler fraction. Then, break down the simpler fraction using a method called partial fractions. Finally, integrate each part separately to get the answer.

🎯 Exam Tip: Always check the degrees of the numerator and denominator in rational functions. If the numerator's degree is equal to or higher than the denominator's, perform polynomial division first before applying partial fractions.

Question 16. Integrate the following function: \( \frac{x-1}{(x+1)(x^2+1)} \)
Answer: We use partial fractions for the given function: \( \frac{x-1}{(x+1)(x^2+1)} = \frac{A}{x+1} + \frac{Bx+C}{x^2+1} \) Multiply both sides by \( (x+1)(x^2+1) \): \( x-1 = A(x^2+1) + (Bx+C)(x+1) \) To find A, set \( x=-1 \): \( (-1)-1 = A((-1)^2+1) \implies -2 = A(1+1) \implies -2 = 2A \implies A=-1 \). Now substitute \( A=-1 \) into the equation: \( x-1 = -1(x^2+1) + (Bx+C)(x+1) \) \( x-1 = -x^2-1 + Bx^2+Bx+Cx+C \) Rearrange terms by powers of \( x \): \( x-1 = (B-1)x^2 + (B+C)x + (C-1) \) Comparing coefficients: Coefficient of \( x^2 \): \( B-1 = 0 \implies B=1 \). Coefficient of \( x \): \( B+C = 1 \implies 1+C = 1 \implies C=0 \). Constant term: \( C-1 = -1 \implies 0-1 = -1 \) (This matches). So, \( A=-1, B=1, C=0 \). The integral becomes: \( \int \left(\frac{-1}{x+1} + \frac{1x+0}{x^2+1}\right) dx = \int \left(\frac{-1}{x+1} + \frac{x}{x^2+1}\right) dx \) \( = -\int \frac{1}{x+1} \,dx + \int \frac{x}{x^2+1} \,dx \) For the second integral, let \( u = x^2+1 \), then \( du = 2x \,dx \implies x \,dx = \frac{1}{2} \,du \). \( = -\log|x+1| + \int \frac{1}{u} \frac{1}{2} \,du \) \( = -\log|x+1| + \frac{1}{2} \log|u| + K \) Substitute back \( u = x^2+1 \): \( = -\log|x+1| + \frac{1}{2} \log|x^2+1| + K \) We can write \( \frac{1}{2} \log|x^2+1| \) as \( \log|\sqrt{x^2+1}| \). \( = \log|\sqrt{x^2+1}| - \log|x+1| + K \) \( = \log \left|\frac{\sqrt{x^2+1}}{x+1}\right| + K \) The final form helps simplify the expression neatly.
In simple words: Break the complex fraction into simpler parts using partial fractions. Then integrate each simple part. Remember to use a substitution for the \( x/(x^2+1) \) term to make integration easier.

🎯 Exam Tip: When using partial fractions for terms like \( \frac{Bx+C}{Ax^2+B'x+C'} \), remember to account for the linear term \( Bx+C \) in the numerator. Also, be careful with the substitution when integrating terms of the form \( \frac{x}{x^2+a^2} \).

Question 17. Integrate the following function: \( \frac{1}{(1+e^x)(1-e^{-x})} \)
Answer: First, rewrite the expression to simplify it for substitution. \( \frac{1}{(1+e^x)(1-e^{-x})} = \frac{1}{(1+e^x)(1-\frac{1}{e^x})} = \frac{1}{(1+e^x)(\frac{e^x-1}{e^x})} = \frac{e^x}{(1+e^x)(e^x-1)} \) Now, let \( t = e^x \). Then \( dt = e^x \,dx \). The integral becomes: \( \int \frac{dt}{(t+1)(t-1)} \) Use partial fractions for \( \frac{1}{(t+1)(t-1)} \): \( \frac{1}{(t+1)(t-1)} = \frac{A}{t+1} + \frac{B}{t-1} \) Multiply by \( (t+1)(t-1) \): \( 1 = A(t-1) + B(t+1) \) To find A, set \( t=-1 \): \( 1 = A(-1-1) \implies 1 = -2A \implies A = -\frac{1}{2} \). To find B, set \( t=1 \): \( 1 = B(1+1) \implies 1 = 2B \implies B = \frac{1}{2} \). So the integral is: \( \int \left(\frac{-\frac{1}{2}}{t+1} + \frac{\frac{1}{2}}{t-1}\right) dt = \frac{1}{2} \int \left(\frac{1}{t-1} - \frac{1}{t+1}\right) dt \) \( = \frac{1}{2} (\log|t-1| - \log|t+1|) + C \) Using logarithm properties, \( \log a - \log b = \log(a/b) \): \( = \frac{1}{2} \log \left|\frac{t-1}{t+1}\right| + C \) Substitute back \( t = e^x \): \( = \frac{1}{2} \log \left|\frac{e^x-1}{e^x+1}\right| + C \) This method transforms a tricky exponential integral into a simpler rational function.
In simple words: First, rewrite the fraction to make it simpler. Then, use a substitution for \( e^x \) to turn it into a fraction with simple \( t \) terms. Break this into two fractions and integrate them separately, then put \( e^x \) back in.

🎯 Exam Tip: For integrals involving \( e^x \) and \( e^{-x} \), try substituting \( t=e^x \) or \( t=e^{-x} \) to convert them into rational functions, which can often be solved using partial fractions. Always simplify the original expression before substitution if possible.

Question 18. Integrate the following function: \( \frac{1}{(e^x-1)^2} \)
Answer: Let \( t = e^x \). Then \( dt = e^x \,dx \). From this, \( dx = \frac{dt}{e^x} = \frac{dt}{t} \). Substitute these into the integral: \( \int \frac{1}{(t-1)^2} \cdot \frac{dt}{t} = \int \frac{1}{t(t-1)^2} \,dt \) Now, use partial fractions for \( \frac{1}{t(t-1)^2} \): \( \frac{1}{t(t-1)^2} = \frac{A}{t} + \frac{B}{t-1} + \frac{C}{(t-1)^2} \) Multiply by \( t(t-1)^2 \): \( 1 = A(t-1)^2 + B t(t-1) + C t \) To find A, set \( t=0 \): \( 1 = A(0-1)^2 \implies 1 = A(1) \implies A=1 \). To find C, set \( t=1 \): \( 1 = C(1) \implies C=1 \). To find B, we can compare coefficients. Expand the equation: \( 1 = A(t^2-2t+1) + B(t^2-t) + Ct \) \( 1 = At^2-2At+A + Bt^2-Bt + Ct \) Group by powers of \( t \): \( 1 = (A+B)t^2 + (-2A-B+C)t + A \) Compare coefficient of \( t^2 \) on both sides (left side has 0): \( A+B = 0 \). Since \( A=1 \), \( 1+B=0 \implies B=-1 \). (Check constant term: \( A=1 \), which matches). So, \( A=1, B=-1, C=1 \). The integral becomes: \( \int \left(\frac{1}{t} - \frac{1}{t-1} + \frac{1}{(t-1)^2}\right) dt \) \( = \int \frac{1}{t} \,dt - \int \frac{1}{t-1} \,dt + \int (t-1)^{-2} \,dt \) \( = \log|t| - \log|t-1| + \frac{(t-1)^{-1}}{-1} + K \) \( = \log|t| - \log|t-1| - \frac{1}{t-1} + K \) Using logarithm properties: \( = \log \left|\frac{t}{t-1}\right| - \frac{1}{t-1} + K \) Substitute back \( t = e^x \): \( = \log \left|\frac{e^x}{e^x-1}\right| - \frac{1}{e^x-1} + K \) This problem combines substitution and partial fractions effectively.
In simple words: Use \( t=e^x \) to change the integral into one involving \( t \). Then, break this new fraction into simpler parts using partial fractions. Integrate each simple part, and finally, change \( t \) back to \( e^x \).

🎯 Exam Tip: When the denominator has a repeated factor, like \( (t-1)^2 \), remember to include all powers of that factor in your partial fraction decomposition (e.g., \( \frac{B}{t-1} \) and \( \frac{C}{(t-1)^2} \)). This is a common point where errors occur.

Question 19. Integrate the following function: \( \frac{e^x}{e^{2x}+5e^x+6} \)
Answer: Let \( t = e^x \). Then \( dt = e^x \,dx \). The integral becomes: \( \int \frac{dt}{t^2+5t+6} \) Factor the denominator: \( t^2+5t+6 = (t+3)(t+2) \). So we have \( \int \frac{dt}{(t+3)(t+2)} \). Use partial fractions for \( \frac{1}{(t+3)(t+2)} \): \( \frac{1}{(t+3)(t+2)} = \frac{A}{t+3} + \frac{B}{t+2} \) Multiply by \( (t+3)(t+2) \): \( 1 = A(t+2) + B(t+3) \) To find A, set \( t=-3 \): \( 1 = A(-3+2) \implies 1 = -A \implies A=-1 \). To find B, set \( t=-2 \): \( 1 = B(-2+3) \implies 1 = B \). So, \( A=-1, B=1 \). The integral becomes: \( \int \left(\frac{-1}{t+3} + \frac{1}{t+2}\right) dt = \int \frac{1}{t+2} \,dt - \int \frac{1}{t+3} \,dt \) \( = \log|t+2| - \log|t+3| + C \) Using logarithm properties, \( \log a - \log b = \log(a/b) \): \( = \log \left|\frac{t+2}{t+3}\right| + C \) Substitute back \( t = e^x \): \( = \log \left|\frac{e^x+2}{e^x+3}\right| + C \) This method transforms exponential functions into more manageable rational forms.
In simple words: Change \( e^x \) to \( t \) to simplify the fraction. Then, split the new fraction using partial fractions. Integrate each part and then substitute \( e^x \) back for \( t \).

🎯 Exam Tip: Factoring the denominator correctly is key for partial fractions. Always double-check your factorization before proceeding with finding the constants A, B, etc.

Question 20. Integrate the following function: \( \frac{\sec^2 x}{(2+\tan x)(3+\tan x)} \)
Answer: Let \( t = \tan x \). Then, the derivative of \( \tan x \) is \( \sec^2 x \), so \( dt = \sec^2 x \,dx \). The integral becomes: \( \int \frac{dt}{(2+t)(3+t)} \) Use partial fractions for \( \frac{1}{(2+t)(3+t)} \): \( \frac{1}{(2+t)(3+t)} = \frac{A}{2+t} + \frac{B}{3+t} \) Multiply by \( (2+t)(3+t) \): \( 1 = A(3+t) + B(2+t) \) To find A, set \( t=-2 \): \( 1 = A(3-2) \implies 1 = A \). To find B, set \( t=-3 \): \( 1 = B(2-3) \implies 1 = -B \implies B=-1 \). So, \( A=1, B=-1 \). The integral becomes: \( \int \left(\frac{1}{2+t} - \frac{1}{3+t}\right) dt \) \( = \int \frac{1}{2+t} \,dt - \int \frac{1}{3+t} \,dt \) \( = \log|2+t| - \log|3+t| + C \) Using logarithm properties: \( = \log \left|\frac{2+t}{3+t}\right| + C \) Substitute back \( t = \tan x \): \( = \log \left|\frac{2+\tan x}{3+\tan x}\right| + C \) This technique effectively transforms trigonometric integrals into algebraic ones for easier solving.
In simple words: Replace \( \tan x \) with \( t \) and \( \sec^2 x \,dx \) with \( dt \) to simplify the problem. Then, use partial fractions to break the new fraction into two simpler parts. Integrate each part, and finally, put \( \tan x \) back into the answer.

🎯 Exam Tip: Always look for suitable substitutions in trigonometric integrals. If you see a function and its derivative (like \( \tan x \) and \( \sec^2 x \)), a substitution is usually the first step to simplify the problem.

Question 21. Integrate the following function: \( \frac{1}{x(x^5+1)} \)
Answer: To simplify this integral, multiply the numerator and denominator by \( x^4 \): \( \int \frac{1}{x(x^5+1)} \,dx = \int \frac{x^4}{x^5(x^5+1)} \,dx \) Now, let \( t = x^5 \). Then \( dt = 5x^4 \,dx \), which means \( x^4 \,dx = \frac{1}{5} \,dt \). Substitute these into the integral: \( \int \frac{1}{t(t+1)} \cdot \frac{1}{5} \,dt = \frac{1}{5} \int \frac{1}{t(t+1)} \,dt \) Use partial fractions for \( \frac{1}{t(t+1)} \): \( \frac{1}{t(t+1)} = \frac{A}{t} + \frac{B}{t+1} \) Multiply by \( t(t+1) \): \( 1 = A(t+1) + Bt \) To find A, set \( t=0 \): \( 1 = A(0+1) \implies A=1 \). To find B, set \( t=-1 \): \( 1 = B(-1) \implies B=-1 \). So, \( A=1, B=-1 \). The integral becomes: \( \frac{1}{5} \int \left(\frac{1}{t} - \frac{1}{t+1}\right) dt \) \( = \frac{1}{5} (\log|t| - \log|t+1|) + C \) Using logarithm properties: \( = \frac{1}{5} \log \left|\frac{t}{t+1}\right| + C \) Substitute back \( t = x^5 \): \( = \frac{1}{5} \log \left|\frac{x^5}{x^5+1}\right| + C \) This problem illustrates how a simple algebraic manipulation can enable effective substitution.
In simple words: First, multiply the top and bottom by \( x^4 \). Then, replace \( x^5 \) with \( t \) to make the fraction simpler. Break it into two simpler fractions, integrate, and then put \( x^5 \) back in place of \( t \).

🎯 Exam Tip: For integrals of the form \( \frac{1}{x(ax^n+b)} \), a useful trick is to multiply the numerator and denominator by \( x^{n-1} \) to facilitate the substitution \( t=x^n \). This converts the integral into a rational function solvable by partial fractions.

Question 22. Integrate the following function: \( \frac{1}{x(a+bx^n)} \)
Answer: To solve this integral, we first perform a substitution. Let \( t = x^n \). Then \( dt = nx^{n-1} \,dx \). From this, \( dx = \frac{dt}{nx^{n-1}} \). Now, rewrite the integrand in terms of \( t \): \( \frac{1}{x(a+bx^n)} \,dx = \frac{1}{x(a+bt)} \cdot \frac{dt}{nx^{n-1}} = \frac{1}{n \cdot x^n (a+bt)} \,dt \) Since \( x^n = t \), we substitute this back: \( = \frac{1}{nt(a+bt)} \,dt \) So the integral becomes \( I = \int \frac{1}{nt(a+bt)} \,dt = \frac{1}{n} \int \frac{1}{t(a+bt)} \,dt \). Now, use partial fractions for \( \frac{1}{t(a+bt)} \): \( \frac{1}{t(a+bt)} = \frac{A}{t} + \frac{B}{a+bt} \) Multiply by \( t(a+bt) \): \( 1 = A(a+bt) + Bt \) To find A, set \( t=0 \): \( 1 = A(a+b(0)) \implies 1 = Aa \implies A=\frac{1}{a} \). To find B, set \( a+bt=0 \implies t = -\frac{a}{b} \): \( 1 = B\left(-\frac{a}{b}\right) \implies B = -\frac{b}{a} \). So, \( A=\frac{1}{a}, B=-\frac{b}{a} \). The integral becomes: \( \frac{1}{n} \int \left(\frac{\frac{1}{a}}{t} + \frac{-\frac{b}{a}}{a+bt}\right) dt = \frac{1}{n} \cdot \frac{1}{a} \int \left(\frac{1}{t} - \frac{b}{a+bt}\right) dt \) \( = \frac{1}{na} \left(\int \frac{1}{t} \,dt - \int \frac{b}{a+bt} \,dt\right) \) \( = \frac{1}{na} (\log|t| - \log|a+bt|) + C \) Using logarithm properties: \( = \frac{1}{na} \log \left|\frac{t}{a+bt}\right| + C \) Substitute back \( t = x^n \): \( = \frac{1}{na} \log \left|\frac{x^n}{a+bx^n}\right| + C \) This problem is a general form and is useful for understanding a class of integrals.
In simple words: Replace \( x^n \) with \( t \) to simplify the fraction. Then, use partial fractions to break the new fraction into two simpler parts. Integrate each part, and finally, put \( x^n \) back in place of \( t \).

🎯 Exam Tip: When dealing with integrals involving \( x \) raised to some power and a term like \( (a+bx^n) \), the substitution \( t=x^n \) is a common and effective strategy. Always ensure to correctly adjust \( dx \) to \( dt \) in the substitution.

Question 23. Integrate the following function: \( \frac{8}{(x+2)(x^2+4)} \)
Answer: We use partial fractions for the given function: \( \frac{8}{(x+2)(x^2+4)} = \frac{A}{x+2} + \frac{Bx+C}{x^2+4} \) Multiply both sides by \( (x+2)(x^2+4) \): \( 8 = A(x^2+4) + (Bx+C)(x+2) \) To find A, set \( x=-2 \): \( 8 = A((-2)^2+4) \implies 8 = A(4+4) \implies 8 = 8A \implies A=1 \). Now substitute \( A=1 \) into the equation: \( 8 = 1(x^2+4) + (Bx+C)(x+2) \) \( 8 = x^2+4 + Bx^2+2Bx+Cx+2C \) Rearrange terms by powers of \( x \): \( 8 = (1+B)x^2 + (2B+C)x + (4+2C) \) 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: \( 4+2C = 8 \implies 4+2(2)=8 \implies 4+4=8 \) (This matches). So, \( A=1, B=-1, C=2 \). The integral becomes: \( \int \left(\frac{1}{x+2} + \frac{-x+2}{x^2+4}\right) dx = \int \left(\frac{1}{x+2} - \frac{x}{x^2+4} + \frac{2}{x^2+4}\right) dx \) \( = \int \frac{1}{x+2} \,dx - \int \frac{x}{x^2+4} \,dx + \int \frac{2}{x^2+4} \,dx \) For the second integral, let \( u = x^2+4 \), then \( du = 2x \,dx \implies x \,dx = \frac{1}{2} \,du \). \( = \log|x+2| - \int \frac{1}{u} \frac{1}{2} \,du + 2 \int \frac{1}{x^2+2^2} \,dx \) \( = \log|x+2| - \frac{1}{2} \log|u| + 2 \cdot \frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right) + K \) Substitute back \( u = x^2+4 \): \( = \log|x+2| - \frac{1}{2} \log|x^2+4| + \tan^{-1}\left(\frac{x}{2}\right) + K \) This solution correctly handles both linear and irreducible quadratic factors in partial fractions.
In simple words: Break the complex fraction into simpler parts using partial fractions, one for \( x+2 \) and another for \( x^2+4 \). Then, integrate each simple part. The \( x^2+4 \) term will involve both a logarithm and an inverse tangent function.

🎯 Exam Tip: When using partial fractions with an irreducible quadratic factor like \( x^2+a^2 \), the numerator for that term should be in the form \( Bx+C \). Remember the standard integral formulas for \( \int \frac{1}{x^2+a^2} \,dx = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) \) and \( \int \frac{x}{x^2+a^2} \,dx = \frac{1}{2}\log|x^2+a^2| \).

Question 24. Integrate the following function: \( \frac{1-\cos x}{\cos x (1+\cos x)} \)
Answer: We use partial fractions for the given trigonometric function, treating \( \cos x \) as a variable for the decomposition: Let \( u = \cos x \). Then the expression is \( \frac{1-u}{u(1+u)} \). \( \frac{1-u}{u(1+u)} = \frac{A}{u} + \frac{B}{1+u} \) Multiply both sides by \( u(1+u) \): \( 1-u = A(1+u) + Bu \) To find A, set \( u=0 \): \( 1-0 = A(1+0) \implies 1 = A \). To find B, set \( 1+u=0 \implies u=-1 \): \( 1-(-1) = B(-1) \implies 2 = -B \implies B=-2 \). So, \( A=1, B=-2 \). The integral becomes: \( \int \left(\frac{1}{\cos x} - \frac{2}{1+\cos x}\right) dx \) \( = \int \sec x \,dx - 2 \int \frac{1}{1+\cos x} \,dx \) We know the integral of \( \sec x \) is \( \log|\sec x + \tan x| \). For the second integral, we use the half-angle identity \( 1+\cos x = 2\cos^2(x/2) \): \( \int \frac{1}{1+\cos x} \,dx = \int \frac{1}{2\cos^2(x/2)} \,dx = \frac{1}{2} \int \sec^2(x/2) \,dx \) Let \( y = x/2 \), then \( dy = \frac{1}{2} \,dx \implies dx = 2 \,dy \). \( = \frac{1}{2} \int \sec^2(y) (2 \,dy) = \int \sec^2(y) \,dy = \tan y + C_1 = \tan(x/2) + C_1 \). Combining both parts: \( = \log|\sec x + \tan x| - 2(\tan(x/2)) + C \) This method cleverly uses partial fractions with trigonometric functions and half-angle identities.
In simple words: First, treat \( \cos x \) like a variable to break the fraction into two simpler parts using partial fractions. Then, integrate the \( \frac{1}{\cos x} \) part as \( \sec x \). For the \( \frac{1}{1+\cos x} \) part, use a special trigonometric rule to change it into \( \sec^2(x/2) \) and then integrate. Combine the results.

🎯 Exam Tip: When integrating rational expressions involving trigonometric functions, sometimes treating the trigonometric function (like \( \sin x \) or \( \cos x \)) as a temporary variable for partial fractions can simplify the decomposition. Also, remember key trigonometric identities like \( 1+\cos x = 2\cos^2(x/2) \) for simplification.