GSEB Class 12 Maths Solutions Chapter 7 Integrals Exercise 7.5

Integrate the following functions:

Question 1. \( \frac{x}{(x+1)(x+2)} \)
Answer:To solve this, we write \( \frac{x}{(x+1)(x+2)} \) as partial fractions: \( \frac{x}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2} \) This implies \( x = A(x+2) + B(x+1) \). Put \( x = -1 \): \( -1 = A(-1+2) + B(0) \) \( -1 = A(1) \) So, \( A = -1 \). Put \( x = -2 \): \( -2 = A(0) + B(-2+1) \) \( -2 = B(-1) \) So, \( B = 2 \). Thus, we can write: \( \frac{x}{(x+1)(x+2)} = \frac{-1}{x+1} + \frac{2}{x+2} \) Now, we integrate both sides: \( \int \frac{x}{(x+1)(x+2)} dx = \int \frac{-1}{x+1} dx + \int \frac{2}{x+2} dx \) \( = -\int \frac{1}{x+1} dx + 2 \int \frac{1}{x+2} dx \) \( = -\log|x+1| + 2\log|x+2| + C \) \( = \log|(x+2)^2| - \log|x+1| + C \) \( = \log \left| \frac{(x+2)^2}{x+1} \right| + C \)
In simple words: We break the complex fraction into simpler parts using 'partial fractions'. Then, we integrate each simple part separately and combine the results, remembering to add the constant 'C' at the end.

Exam Tip: For partial fraction decomposition, always ensure the degree of the numerator is less than the degree of the denominator. If it's not, perform polynomial long division first.

Question 2. \( \frac{1}{x^{2}-9} \)
Answer:To integrate \( \frac{1}{x^{2}-9} \), we first factor the denominator: \( x^2 - 9 = (x-3)(x+3) \). We express the fraction using partial fractions: \( \frac{1}{x^{2}-9} = \frac{1}{(x-3)(x+3)} = \frac{A}{x-3} + \frac{B}{x+3} \) This implies \( 1 = A(x+3) + B(x-3) \) (Equation 1). Put \( x = 3 \) in Equation 1: \( 1 = A(3+3) + B(0) \) \( 1 = 6A \) So, \( A = \frac{1}{6} \). Put \( x = -3 \) in Equation 1: \( 1 = A(0) + B(-3-3) \) \( 1 = -6B \) So, \( B = -\frac{1}{6} \). Thus, we can write: \( \frac{1}{x^{2}-9} = \frac{\frac{1}{6}}{x-3} + \frac{-\frac{1}{6}}{x+3} = \frac{1}{6} \left( \frac{1}{x-3} - \frac{1}{x+3} \right) \) Now, we integrate both sides: \( \int \frac{1}{x^{2}-9} dx = \int \frac{1}{6} \left( \frac{1}{x-3} - \frac{1}{x+3} \right) dx \) \( = \frac{1}{6} \left( \int \frac{1}{x-3} dx - \int \frac{1}{x+3} dx \right) \) \( = \frac{1}{6} (\log|x-3| - \log|x+3|) + C \) \( = \frac{1}{6} \log \left| \frac{x-3}{x+3} \right| + C \)
In simple words: First, break the fraction into two simpler parts. Then, integrate each part separately using the logarithm rule for fractions. Finally, combine them and remember the integration constant.

Exam Tip: Recognize the standard integral form \( \int \frac{1}{x^2 - a^2} dx = \frac{1}{2a} \log \left| \frac{x-a}{x+a} \right| + C \). Here, \( a=3 \). Applying this direct formula saves time.

Question 3. \( \frac{3x-1}{(x-1)(x-2)(x-3)} \)
Answer:To integrate \( \frac{3x-1}{(x-1)(x-2)(x-3)} \), we use partial fraction decomposition: \( \frac{3x-1}{(x-1)(x-2)(x-3)} = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3} \) This implies \( 3x-1 = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2) \) (Equation 1). Put \( x = 1 \) in Equation 1: \( 3(1)-1 = A(1-2)(1-3) + B(0) + C(0) \) \( 2 = A(-1)(-2) \) \( 2 = 2A \) So, \( A = 1 \). Put \( x = 2 \) in Equation 1: \( 3(2)-1 = A(0) + B(2-1)(2-3) + C(0) \) \( 5 = B(1)(-1) \) \( 5 = -B \) So, \( B = -5 \). Put \( x = 3 \) in Equation 1: \( 3(3)-1 = A(0) + B(0) + C(3-1)(3-2) \) \( 8 = C(2)(1) \) \( 8 = 2C \) So, \( C = 4 \). Thus, we can write: \( \frac{3x-1}{(x-1)(x-2)(x-3)} = \frac{1}{x-1} - \frac{5}{x-2} + \frac{4}{x-3} \) Now, we integrate both sides: \( \int \frac{3x-1}{(x-1)(x-2)(x-3)} dx = \int \frac{1}{x-1} dx - \int \frac{5}{x-2} dx + \int \frac{4}{x-3} dx \) \( = \int \frac{1}{x-1} dx - 5 \int \frac{1}{x-2} dx + 4 \int \frac{1}{x-3} dx \) \( = \log|x-1| - 5\log|x-2| + 4\log|x-3| + C \)
In simple words: Break the fraction into three simpler parts. Then, integrate each simple part using the log rule. Combine all the results at the end with the constant C.

Exam Tip: When finding coefficients for partial fractions with distinct linear factors, the cover-up method (Heaviside's method) is very efficient. For example, to find A, cover (x-1) and substitute x=1 into the rest of the expression.

Question 4. \( \frac{x}{(x-1)(x-2)(x-3)} \)
Answer:To integrate \( \frac{x}{(x-1)(x-2)(x-3)} \), we use partial fraction decomposition: \( \frac{x}{(x-1)(x-2)(x-3)} = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3} \) This implies \( x = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2) \) (Equation 1). Put \( x = 1 \) in Equation 1: \( 1 = A(1-2)(1-3) \) \( 1 = A(-1)(-2) \) \( 1 = 2A \) So, \( A = \frac{1}{2} \). Put \( x = 2 \) in Equation 1: \( 2 = B(2-1)(2-3) \) \( 2 = B(1)(-1) \) \( 2 = -B \) So, \( B = -2 \). Put \( x = 3 \) in Equation 1: \( 3 = C(3-1)(3-2) \) \( 3 = C(2)(1) \) \( 3 = 2C \) So, \( C = \frac{3}{2} \). Thus, we can write: \( \frac{x}{(x-1)(x-2)(x-3)} = \frac{\frac{1}{2}}{x-1} - \frac{2}{x-2} + \frac{\frac{3}{2}}{x-3} \) Now, we integrate both sides: \( \int \frac{x}{(x-1)(x-2)(x-3)} dx = \int \frac{1}{2(x-1)} dx - \int \frac{2}{x-2} dx + \int \frac{3}{2(x-3)} dx \) \( = \frac{1}{2} \int \frac{1}{x-1} dx - 2 \int \frac{1}{x-2} dx + \frac{3}{2} \int \frac{1}{x-3} dx \) \( = \frac{1}{2}\log|x-1| - 2\log|x-2| + \frac{3}{2}\log|x-3| + C \)
In simple words: First, break down the complex fraction into a sum of simpler fractions. Then, integrate each of these simpler parts separately using the logarithmic integration rule. Finally, combine the results and include the constant of integration.

Exam Tip: Pay careful attention to signs when calculating the coefficients A, B, and C. A small error in a sign can lead to a completely wrong answer.

Question 5. \( \frac{2 x}{x^{2}+3x+2} \)
Answer:To integrate \( \frac{2x}{x^2+3x+2} \), we factor the denominator: \( x^2+3x+2 = (x+1)(x+2) \). We express the fraction using partial fractions: \( \frac{2x}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2} \) This implies \( 2x = A(x+2) + B(x+1) \) (Equation 1). Put \( x = -1 \) in Equation 1: \( 2(-1) = A(-1+2) + B(0) \) \( -2 = A(1) \) So, \( A = -2 \). Put \( x = -2 \) in Equation 1: \( 2(-2) = A(0) + B(-2+1) \) \( -4 = B(-1) \) So, \( B = 4 \). Thus, we can write: \( \frac{2x}{x^2+3x+2} = \frac{-2}{x+1} + \frac{4}{x+2} \) Now, we integrate both sides: \( \int \frac{2x}{x^2+3x+2} dx = \int \frac{-2}{x+1} dx + \int \frac{4}{x+2} dx \) \( = -2 \int \frac{1}{x+1} dx + 4 \int \frac{1}{x+2} dx \) \( = -2\log|x+1| + 4\log|x+2| + C \)
In simple words: First, break the fraction into simpler parts by factoring the bottom. Then, integrate each simple part using the rule for logarithms. Combine the results and add C.

Exam Tip: Always factor the denominator completely before applying partial fraction decomposition. This ensures all possible linear or irreducible quadratic factors are accounted for.

Question 6. \( \frac{1-x^{2}}{x(1-2 x)} \)
Answer:The given function \( \frac{1-x^2}{x(1-2x)} = \frac{1-x^2}{x-2x^2} \) is an improper fraction because the degree of the numerator (2) is equal to the degree of the denominator (2). Therefore, we must first perform polynomial long division. Dividing \( 1-x^2 \) by \( x-2x^2 \): \[ \frac{-x^2+1}{-2x^2+x} = \frac{1}{2} + \frac{\frac{1}{2}x+1}{-2x^2+x} = \frac{1}{2} + \frac{x-2}{2x-4x^2} \] Let's perform the long division systematically: Multiply \( (-2x^2+x) \) by \( \frac{1}{2} \): \( -x^2 + \frac{1}{2}x \) Subtract this from \( -x^2+1 \): \( (-x^2+1) - (-x^2 + \frac{1}{2}x) = 1 - \frac{1}{2}x \) So, \( \frac{1-x^2}{x(1-2x)} = \frac{1}{2} + \frac{1 - \frac{1}{2}x}{x(1-2x)} = \frac{1}{2} + \frac{2-x}{2x(1-2x)} \) Now, we need to decompose \( \frac{2-x}{2x(1-2x)} \) using partial fractions. Let \( \frac{2-x}{x(1-2x)} = \frac{A}{x} + \frac{B}{1-2x} \) (ignoring the factor of 2 for a moment, we'll put it back later) This implies \( 2-x = A(1-2x) + Bx \) (Equation 2). Put \( x = 0 \) in Equation 2: \( 2-0 = A(1-0) + B(0) \) \( 2 = A(1) \) So, \( A = 2 \). Put \( x = \frac{1}{2} \) in Equation 2: \( 2 - \frac{1}{2} = A(0) + B(\frac{1}{2}) \) \( \frac{3}{2} = \frac{1}{2}B \) So, \( B = 3 \). Thus, \( \frac{2-x}{x(1-2x)} = \frac{2}{x} + \frac{3}{1-2x} \). Putting it back into the original expression: \( \frac{1-x^2}{x(1-2x)} = \frac{1}{2} + \frac{1}{2} \left( \frac{2}{x} + \frac{3}{1-2x} \right) = \frac{1}{2} + \frac{1}{x} + \frac{3}{2(1-2x)} \) Now, we integrate both sides: \( \int \frac{1-x^2}{x(1-2x)} dx = \int \frac{1}{2} dx + \int \frac{1}{x} dx + \int \frac{3}{2(1-2x)} dx \) \( = \frac{1}{2}x + \log|x| + \frac{3}{2} \int \frac{1}{1-2x} dx \) For \( \int \frac{1}{1-2x} dx \), let \( u = 1-2x \), so \( du = -2dx \), meaning \( dx = -\frac{1}{2}du \). \( \int \frac{1}{1-2x} dx = \int \frac{1}{u} (-\frac{1}{2}) du = -\frac{1}{2} \log|u| = -\frac{1}{2} \log|1-2x| \) So, the full integral is: \( = \frac{1}{2}x + \log|x| + \frac{3}{2} \left( -\frac{1}{2} \log|1-2x| \right) + C \) \( = \frac{1}{2}x + \log|x| - \frac{3}{4} \log|1-2x| + C \)
In simple words: Since the top and bottom parts have the same highest power, first divide them. Then, break the remaining fraction into simpler pieces. Integrate each piece, remembering the rule for logarithms and adjusting for any constants or negative signs from the 'x' term.

Exam Tip: When the integrand is an improper fraction (degree of numerator \( \geq \) degree of denominator), always perform polynomial long division first. This creates a polynomial part and a proper rational function, which can then be decomposed using partial fractions.

Question 7. \( \frac{x}{\left(x^{2}+1\right)(x-1)} \)
Answer:To integrate \( \frac{x}{(x^2+1)(x-1)} \), we use partial fraction decomposition. Here we have an irreducible quadratic factor \( (x^2+1) \) and a linear factor \( (x-1) \). \( \frac{x}{(x^2+1)(x-1)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+1} \) This implies \( x = A(x^2+1) + (Bx+C)(x-1) \) (Equation 1). Put \( x = 1 \) in Equation 1: \( 1 = A(1^2+1) + (B(1)+C)(1-1) \) \( 1 = A(2) + 0 \) \( 1 = 2A \) So, \( A = \frac{1}{2} \). Now, expand Equation 1: \( x = A(x^2+1) + Bx^2 - Bx + Cx - C \) \( x = (A+B)x^2 + (-B+C)x + (A-C) \) Compare coefficients of \( x^2 \): \( 0 = A+B \) Since \( A = \frac{1}{2} \), then \( B = -A = -\frac{1}{2} \). Compare constant terms: \( 0 = A-C \) Since \( A = \frac{1}{2} \), then \( C = A = \frac{1}{2} \). Thus, we can write: \( \frac{x}{(x^2+1)(x-1)} = \frac{\frac{1}{2}}{x-1} + \frac{-\frac{1}{2}x+\frac{1}{2}}{x^2+1} = \frac{1}{2(x-1)} - \frac{x}{2(x^2+1)} + \frac{1}{2(x^2+1)} \) Now, we integrate both sides: \( \int \frac{x}{(x^2+1)(x-1)} dx = \int \frac{1}{2(x-1)} dx - \int \frac{x}{2(x^2+1)} dx + \int \frac{1}{2(x^2+1)} dx \) \( = \frac{1}{2} \int \frac{1}{x-1} dx - \frac{1}{2} \int \frac{x}{x^2+1} dx + \frac{1}{2} \int \frac{1}{x^2+1} dx \) For \( \int \frac{x}{x^2+1} dx \), let \( u = x^2+1 \), so \( du = 2x dx \), meaning \( x dx = \frac{1}{2}du \). \( \int \frac{x}{x^2+1} dx = \int \frac{1}{u} \frac{1}{2} du = \frac{1}{2} \log|u| = \frac{1}{2} \log|x^2+1| \) And \( \int \frac{1}{x^2+1} dx = \tan^{-1}x \). So, the full integral is: \( = \frac{1}{2}\log|x-1| - \frac{1}{2} \left( \frac{1}{2} \log|x^2+1| \right) + \frac{1}{2}\tan^{-1}x + C \) \( = \frac{1}{2}\log|x-1| - \frac{1}{4}\log|x^2+1| + \frac{1}{2}\tan^{-1}x + C \)
In simple words: Break the fraction into simpler parts, one for the linear factor and one for the quadratic factor. Then, integrate each part. Remember to use a 'u-substitution' for the x/(x^2+1) term and the inverse tangent for 1/(x^2+1).

Exam Tip: When dealing with irreducible quadratic factors like \( x^2+1 \), the numerator in the partial fraction should be of the form \( Bx+C \). Remember to compare coefficients for different powers of x (or use specific x values) to solve for A, B, and C.

Question 8. \( \frac{x}{(x-1)^{2}(x+2)} \)
Answer:To integrate \( \frac{x}{(x-1)^2(x+2)} \), we use partial fraction decomposition. Here we have a repeated linear factor \( (x-1)^2 \) and a distinct linear factor \( (x+2) \). \( \frac{x}{(x-1)^2(x+2)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+2} \) This implies \( x = A(x-1)(x+2) + B(x+2) + C(x-1)^2 \) (Equation 1). Put \( x = 1 \) in Equation 1: \( 1 = A(0) + B(1+2) + C(0) \) \( 1 = B(3) \) So, \( B = \frac{1}{3} \). Put \( x = -2 \) in Equation 1: \( -2 = A(0) + B(0) + C(-2-1)^2 \) \( -2 = C(-3)^2 \) \( -2 = 9C \) So, \( C = -\frac{2}{9} \). To find \( A \), compare the coefficients of \( x^2 \) on both sides of Equation 1: \( x = A(x^2+x-2) + B(x+2) + C(x^2-2x+1) \) \( x = Ax^2+Ax-2A + Bx+2B + Cx^2-2Cx+C \) \( x = (A+C)x^2 + (A+B-2C)x + (-2A+2B+C) \) Coefficient of \( x^2 \): \( 0 = A+C \) Since \( C = -\frac{2}{9} \), then \( A = -C = -(-\frac{2}{9}) = \frac{2}{9} \). Thus, we can write: \( \frac{x}{(x-1)^2(x+2)} = \frac{\frac{2}{9}}{x-1} + \frac{\frac{1}{3}}{(x-1)^2} - \frac{\frac{2}{9}}{x+2} \) Now, we integrate both sides: \( \int \frac{x}{(x-1)^2(x+2)} dx = \frac{2}{9} \int \frac{1}{x-1} dx + \frac{1}{3} \int \frac{1}{(x-1)^2} dx - \frac{2}{9} \int \frac{1}{x+2} dx \) \( = \frac{2}{9}\log|x-1| + \frac{1}{3} \left( -\frac{1}{x-1} \right) - \frac{2}{9}\log|x+2| + C \) \( = \frac{2}{9}\log|x-1| - \frac{1}{3(x-1)} - \frac{2}{9}\log|x+2| + C \) \( = \frac{2}{9} \left( \log|x-1| - \log|x+2| \right) - \frac{1}{3(x-1)} + C \) \( = \frac{2}{9}\log \left| \frac{x-1}{x+2} \right| - \frac{1}{3(x-1)} + C \)
In simple words: Break the fraction into three simpler parts because one factor is repeated. Find the values for A, B, and C. Then, integrate each part separately, remembering that the integral of \( \frac{1}{(x-a)^2} \) is \( -\frac{1}{x-a} \).

Exam Tip: When dealing with repeated linear factors like \( (x-a)^n \), include terms \( \frac{A_1}{x-a}, \frac{A_2}{(x-a)^2}, \dots, \frac{A_n}{(x-a)^n} \) in the partial fraction decomposition. The integral of \( \frac{1}{(x-a)^n} \) is \( \frac{(x-a)^{-n+1}}{-n+1} \) for \( n \neq 1 \).

Question 9. \( \frac{3 x+5}{x^{3}-x^{2}-x+1} \)
Answer:First, factor the denominator: \( x^3-x^2-x+1 \). \( x^3-x^2-x+1 = x^2(x-1) - 1(x-1) = (x^2-1)(x-1) = (x-1)(x+1)(x-1) = (x-1)^2(x+1) \). Now, we use partial fraction decomposition: \( \frac{3x+5}{(x-1)^2(x+1)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+1} \) This implies \( 3x+5 = A(x-1)(x+1) + B(x+1) + C(x-1)^2 \) (Equation 1). Put \( x = 1 \) in Equation 1: \( 3(1)+5 = A(0) + B(1+1) + C(0) \) \( 8 = 2B \) So, \( B = 4 \). Put \( x = -1 \) in Equation 1: \( 3(-1)+5 = A(0) + B(0) + C(-1-1)^2 \) \( -3+5 = C(-2)^2 \) \( 2 = 4C \) So, \( C = \frac{2}{4} = \frac{1}{2} \). To find \( A \), compare the coefficients of \( x^2 \) on both sides of Equation 1: \( 3x+5 = A(x^2-1) + B(x+1) + C(x^2-2x+1) \) \( 3x+5 = Ax^2-A + Bx+B + Cx^2-2Cx+C \) \( 3x+5 = (A+C)x^2 + (B-2C)x + (-A+B+C) \) Coefficient of \( x^2 \): \( 0 = A+C \) Since \( C = \frac{1}{2} \), then \( A = -C = -\frac{1}{2} \). Thus, we can write: \( \frac{3x+5}{(x-1)^2(x+1)} = \frac{-\frac{1}{2}}{x-1} + \frac{4}{(x-1)^2} + \frac{\frac{1}{2}}{x+1} \) Now, we integrate both sides: \( \int \frac{3x+5}{(x-1)^2(x+1)} dx = -\frac{1}{2} \int \frac{1}{x-1} dx + 4 \int \frac{1}{(x-1)^2} dx + \frac{1}{2} \int \frac{1}{x+1} dx \) \( = -\frac{1}{2}\log|x-1| + 4 \left( -\frac{1}{x-1} \right) + \frac{1}{2}\log|x+1| + C \) \( = \frac{1}{2}\log|x+1| - \frac{1}{2}\log|x-1| - \frac{4}{x-1} + C \) \( = \frac{1}{2}\log \left| \frac{x+1}{x-1} \right| - \frac{4}{x-1} + C \)
In simple words: First, factor the bottom part of the fraction. Then, split the fraction into simpler parts using the rules for repeated factors. Find the missing numbers (A, B, C). Finally, integrate each simpler part, remembering the special rule for repeated factors and adding the constant C.

Exam Tip: Factoring the denominator is the crucial first step. If the denominator has a repeated factor, ensure you include terms for each power of that factor in the partial fraction setup.

Question 10. \( \frac{2 x-3}{\left(x^{2}-1\right)(2 x+3)} \)
Answer:First, factor the denominator: \( x^2-1 = (x-1)(x+1) \). So, \( \frac{2x-3}{(x^2-1)(2x+3)} = \frac{2x-3}{(x-1)(x+1)(2x+3)} \). Now, we use partial fraction decomposition: \( \frac{2x-3}{(x-1)(x+1)(2x+3)} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{C}{2x+3} \) This implies \( 2x-3 = A(x+1)(2x+3) + B(x-1)(2x+3) + C(x-1)(x+1) \) (Equation 1). Put \( x = 1 \) in Equation 1: \( 2(1)-3 = A(1+1)(2(1)+3) + B(0) + C(0) \) \( -1 = A(2)(5) \) \( -1 = 10A \) So, \( A = -\frac{1}{10} \). Put \( x = -1 \) in Equation 1: \( 2(-1)-3 = A(0) + B(-1-1)(2(-1)+3) + C(0) \) \( -5 = B(-2)(-2+3) \) \( -5 = B(-2)(1) \) \( -5 = -2B \) So, \( B = \frac{5}{2} \). Put \( x = -\frac{3}{2} \) in Equation 1: \( 2(-\frac{3}{2})-3 = A(0) + B(0) + C(-\frac{3}{2}-1)(-\frac{3}{2}+1) \) \( -3-3 = C(-\frac{5}{2})(-\frac{1}{2}) \) \( -6 = C(\frac{5}{4}) \) So, \( C = -6 \times \frac{4}{5} = -\frac{24}{5} \). Thus, we can write: \( \frac{2x-3}{(x-1)(x+1)(2x+3)} = \frac{-\frac{1}{10}}{x-1} + \frac{\frac{5}{2}}{x+1} + \frac{-\frac{24}{5}}{2x+3} \) Now, we integrate both sides: \( \int \frac{2x-3}{(x-1)(x+1)(2x+3)} dx = -\frac{1}{10} \int \frac{1}{x-1} dx + \frac{5}{2} \int \frac{1}{x+1} dx - \frac{24}{5} \int \frac{1}{2x+3} dx \) \( = -\frac{1}{10}\log|x-1| + \frac{5}{2}\log|x+1| - \frac{24}{5} \left( \frac{1}{2}\log|2x+3| \right) + C \) \( = -\frac{1}{10}\log|x-1| + \frac{5}{2}\log|x+1| - \frac{12}{5}\log|2x+3| + C \)
In simple words: First, factor the bottom part of the fraction completely. Then, use partial fractions to break it into three simpler parts. Find the unknown numbers (A, B, C) by using specific x values. Finally, integrate each simpler part using the logarithm rule, taking care of any coefficients or terms inside the logarithm.

Exam Tip: Be careful with the integration of terms like \( \frac{1}{ax+b} \); it integrates to \( \frac{1}{a} \log|ax+b| \). Don't forget the factor of \( \frac{1}{a} \) from the derivative of the inner function.

Question 11. \( \frac{5 x}{(x+1)\left(x^{2}-4\right)} \)
Answer:First, factor the denominator: \( x^2-4 = (x-2)(x+2) \). So, \( \frac{5x}{(x+1)(x^2-4)} = \frac{5x}{(x+1)(x+2)(x-2)} \). Now, we use partial fraction decomposition: \( \frac{5x}{(x+1)(x+2)(x-2)} = \frac{A}{x+1} + \frac{B}{x+2} + \frac{C}{x-2} \) This implies \( 5x = A(x+2)(x-2) + B(x+1)(x-2) + C(x+1)(x+2) \) (Equation 1). Put \( x = -1 \) in Equation 1: \( 5(-1) = A(-1+2)(-1-2) + B(0) + C(0) \) \( -5 = A(1)(-3) \) \( -5 = -3A \) So, \( A = \frac{5}{3} \). Put \( x = -2 \) in Equation 1: \( 5(-2) = A(0) + B(-2+1)(-2-2) + C(0) \) \( -10 = B(-1)(-4) \) \( -10 = 4B \) So, \( B = -\frac{10}{4} = -\frac{5}{2} \). Put \( x = 2 \) in Equation 1: \( 5(2) = A(0) + B(0) + C(2+1)(2+2) \) \( 10 = C(3)(4) \) \( 10 = 12C \) So, \( C = \frac{10}{12} = \frac{5}{6} \). Thus, we can write: \( \frac{5x}{(x+1)(x^2-4)} = \frac{\frac{5}{3}}{x+1} + \frac{-\frac{5}{2}}{x+2} + \frac{\frac{5}{6}}{x-2} \) Now, we integrate both sides: \( \int \frac{5x}{(x+1)(x^2-4)} dx = \frac{5}{3} \int \frac{1}{x+1} dx - \frac{5}{2} \int \frac{1}{x+2} dx + \frac{5}{6} \int \frac{1}{x-2} dx \) \( = \frac{5}{3}\log|x+1| - \frac{5}{2}\log|x+2| + \frac{5}{6}\log|x-2| + C \)
In simple words: First, break down the fraction into simpler parts after factoring the bottom. Then, calculate the unknown coefficients. Finally, integrate each simpler part using the logarithm rule and add the constant of integration.

Exam Tip: Be mindful of algebraic signs during calculation. It's often helpful to double-check the coefficients by substituting a fourth arbitrary value of x (e.g., x=0) into the decomposed form and comparing it to the original expression.

Question 12. \( \frac{x^{3}+x+1}{x^{2}-1} \)
Answer:The given function \( \frac{x^3+x+1}{x^2-1} \) is an improper fraction because the degree of the numerator (3) is greater than the degree of the denominator (2). Therefore, we must first perform polynomial long division. Dividing \( x^3+x+1 \) by \( x^2-1 \): \[ \begin{array}{c|cc cc} \multicolumn{2}{r}{x} \\ \cline{2-5} x^2-1 & x^3 & +0x^2 & +x & +1 \\ \multicolumn{2}{r}{x^3} & & -x \\ \cline{2-4} \multicolumn{2}{r}{0} & +0x^2 & +2x & +1 \\ \end{array} \] So, \( \frac{x^3+x+1}{x^2-1} = x + \frac{2x+1}{x^2-1} \). Now, we need to decompose \( \frac{2x+1}{x^2-1} \) using partial fractions. Factor the denominator: \( x^2-1 = (x-1)(x+1) \). Let \( \frac{2x+1}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1} \) This implies \( 2x+1 = A(x+1) + B(x-1) \) (Equation 2). Put \( x = 1 \) in Equation 2: \( 2(1)+1 = A(1+1) + B(0) \) \( 3 = 2A \) So, \( A = \frac{3}{2} \). Put \( x = -1 \) in Equation 2: \( 2(-1)+1 = A(0) + B(-1-1) \) \( -1 = -2B \) So, \( B = \frac{1}{2} \). Thus, \( \frac{2x+1}{x^2-1} = \frac{\frac{3}{2}}{x-1} + \frac{\frac{1}{2}}{x+1} \). Putting it back into the original expression: \( \frac{x^3+x+1}{x^2-1} = x + \frac{3}{2(x-1)} + \frac{1}{2(x+1)} \) Now, we integrate both sides: \( \int \frac{x^3+x+1}{x^2-1} dx = \int x dx + \int \frac{3}{2(x-1)} dx + \int \frac{1}{2(x+1)} dx \) \( = \frac{x^2}{2} + \frac{3}{2} \int \frac{1}{x-1} dx + \frac{1}{2} \int \frac{1}{x+1} dx \) \( = \frac{x^2}{2} + \frac{3}{2}\log|x-1| + \frac{1}{2}\log|x+1| + C \)
In simple words: Since the top part has a higher power than the bottom, first divide them using long division. Then, break the remaining fraction into simpler parts. Integrate the polynomial part and each simpler fraction separately, adding the constant C at the end.

Exam Tip: Always remember to integrate the polynomial part obtained from long division, not just the rational part. The integral of x is \( \frac{x^2}{2} \).

Question 13. \( \frac{2}{(1-x)\left(1+x^{2}\right)} \)
Answer:To integrate \( \frac{2}{(1-x)(1+x^2)} \), we use partial fraction decomposition. Here we have a linear factor \( (1-x) \) and an irreducible quadratic factor \( (1+x^2) \). \( \frac{2}{(1-x)(1+x^2)} = \frac{A}{1-x} + \frac{Bx+C}{1+x^2} \) This implies \( 2 = A(1+x^2) + (Bx+C)(1-x) \) (Equation 1). Put \( x = 1 \) in Equation 1: \( 2 = A(1+1^2) + (B(1)+C)(1-1) \) \( 2 = A(2) + 0 \) So, \( A = 1 \). Now, expand Equation 1: \( 2 = A+Ax^2 + Bx-Bx^2 + C-Cx \) \( 2 = (A-B)x^2 + (B-C)x + (A+C) \) Compare coefficients of \( x^2 \): \( 0 = A-B \) Since \( A = 1 \), then \( B = A = 1 \). Compare constant terms: \( 2 = A+C \) Since \( A = 1 \), then \( 2 = 1+C \), so \( C = 1 \). Thus, we can write: \( \frac{2}{(1-x)(1+x^2)} = \frac{1}{1-x} + \frac{x+1}{1+x^2} = \frac{1}{1-x} + \frac{x}{1+x^2} + \frac{1}{1+x^2} \) Now, we integrate both sides: \( \int \frac{2}{(1-x)(1+x^2)} dx = \int \frac{1}{1-x} dx + \int \frac{x}{1+x^2} dx + \int \frac{1}{1+x^2} dx \) For \( \int \frac{1}{1-x} dx \), let \( u = 1-x \), so \( du = -dx \), meaning \( dx = -du \). \( \int \frac{1}{1-x} dx = \int \frac{1}{u} (-1) du = -\log|u| = -\log|1-x| \) For \( \int \frac{x}{1+x^2} dx \), let \( v = 1+x^2 \), so \( dv = 2x dx \), meaning \( x dx = \frac{1}{2}dv \). \( \int \frac{x}{1+x^2} dx = \int \frac{1}{v} \frac{1}{2} dv = \frac{1}{2}\log|v| = \frac{1}{2}\log|1+x^2| \) And \( \int \frac{1}{1+x^2} dx = \tan^{-1}x \). So, the full integral is: \( = -\log|1-x| + \frac{1}{2}\log|1+x^2| + \tan^{-1}x + C \)
In simple words: Break the fraction into simpler parts, one for the straight factor and one for the curved factor. Find the unknown numbers A, B, and C. Then, integrate each part separately using the logarithm and inverse tangent rules, remembering to adjust for any negative signs in the denominator.

Exam Tip: Be careful with the linear term \( (1-x) \) in the denominator; its integral is \( -\log|1-x| \) because of the \( -1 \) coefficient of x. Also, recognize standard integrals like \( \int \frac{x}{a^2+x^2} dx = \frac{1}{2} \log(a^2+x^2) \) and \( \int \frac{1}{a^2+x^2} dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) \).

Question 14. \( \frac{3 x-1}{(x+2)^{2}} \)
Answer:To integrate \( \frac{3x-1}{(x+2)^2} \), we can use partial fraction decomposition, as there is a repeated linear factor. \( \frac{3x-1}{(x+2)^2} = \frac{A}{x+2} + \frac{B}{(x+2)^2} \) This implies \( 3x-1 = A(x+2) + B \) (Equation 1). Put \( x = -2 \) in Equation 1: \( 3(-2)-1 = A(0) + B \) \( -6-1 = B \) So, \( B = -7 \). Now, to find \( A \), compare the coefficients of \( x \): \( 3x-1 = Ax+2A + B \) Coefficient of \( x \): \( 3 = A \). So, \( A = 3 \). Thus, we can write: \( \frac{3x-1}{(x+2)^2} = \frac{3}{x+2} + \frac{-7}{(x+2)^2} \) Now, we integrate both sides: \( \int \frac{3x-1}{(x+2)^2} dx = \int \frac{3}{x+2} dx - \int \frac{7}{(x+2)^2} dx \) \( = 3 \int \frac{1}{x+2} dx - 7 \int (x+2)^{-2} dx \) \( = 3\log|x+2| - 7 \left( \frac{(x+2)^{-2+1}}{-2+1} \right) + C \) \( = 3\log|x+2| - 7 \left( \frac{(x+2)^{-1}}{-1} \right) + C \) \( = 3\log|x+2| + \frac{7}{x+2} + C \)
In simple words: Break the fraction into two simpler parts, accounting for the repeated factor. Find the values for A and B. Then, integrate each part separately, remembering that the integral of \( \frac{1}{(ax+b)^2} \) is \( -\frac{1}{a(ax+b)} \).

Exam Tip: For expressions like \( \frac{Px+Q}{(ax+b)^2} \), an alternative is to write \( Px+Q = A(ax+b) + B \). This directly gives A and B, simplifying partial fraction decomposition for such forms.

Question 15. \( \frac{1}{x^{4}-1} \)
Answer:First, factor the denominator: \( x^4-1 = (x^2-1)(x^2+1) = (x-1)(x+1)(x^2+1) \). Now, we use partial fraction decomposition: \( \frac{1}{(x-1)(x+1)(x^2+1)} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{Cx+D}{x^2+1} \) This implies \( 1 = A(x+1)(x^2+1) + B(x-1)(x^2+1) + (Cx+D)(x-1)(x+1) \) (Equation 1). Put \( x = 1 \) in Equation 1: \( 1 = A(1+1)(1^2+1) + B(0) + (C(1)+D)(0) \) \( 1 = A(2)(2) \) \( 1 = 4A \) So, \( A = \frac{1}{4} \). Put \( x = -1 \) in Equation 1: \( 1 = A(0) + B(-1-1)((-1)^2+1) + (C(-1)+D)(0) \) \( 1 = B(-2)(1+1) \) \( 1 = B(-2)(2) \) \( 1 = -4B \) So, \( B = -\frac{1}{4} \). Now, expand Equation 1: \( 1 = A(x^3+x+x^2+1) + B(x^3+x-x^2-1) + (Cx+D)(x^2-1) \) \( 1 = A(x^3+x^2+x+1) + B(x^3-x^2+x-1) + Cx^3-Cx + Dx^2-D \) \( 1 = (A+B+C)x^3 + (A-B+D)x^2 + (A+B-C)x + (A-B-D) \) Compare coefficients of \( x^3 \): \( 0 = A+B+C \) \( 0 = \frac{1}{4} - \frac{1}{4} + C \) So, \( C = 0 \). Compare constant terms: \( 1 = A-B-D \) \( 1 = \frac{1}{4} - (-\frac{1}{4}) - D \) \( 1 = \frac{1}{4} + \frac{1}{4} - D \) \( 1 = \frac{2}{4} - D \) \( 1 = \frac{1}{2} - D \) \( D = \frac{1}{2} - 1 = -\frac{1}{2} \). Thus, we can write: \( \frac{1}{x^4-1} = \frac{\frac{1}{4}}{x-1} + \frac{-\frac{1}{4}}{x+1} + \frac{0x-\frac{1}{2}}{x^2+1} = \frac{1}{4(x-1)} - \frac{1}{4(x+1)} - \frac{1}{2(x^2+1)} \) Now, we integrate both sides: \( \int \frac{1}{x^4-1} dx = \frac{1}{4} \int \frac{1}{x-1} dx - \frac{1}{4} \int \frac{1}{x+1} dx - \frac{1}{2} \int \frac{1}{x^2+1} dx \) \( = \frac{1}{4}\log|x-1| - \frac{1}{4}\log|x+1| - \frac{1}{2}\tan^{-1}x + C \) \( = \frac{1}{4} \log \left| \frac{x-1}{x+1} \right| - \frac{1}{2}\tan^{-1}x + C \)
In simple words: First, completely factor the denominator, including the quadratic factor. Then, use partial fractions to break the expression into simpler parts, finding the unknown coefficients. Finally, integrate each simpler part, using logarithms for linear factors and inverse tangent for the quadratic factor.

Exam Tip: Remember that \( x^4-1 \) is a difference of squares twice, leading to \( (x-1)(x+1)(x^2+1) \). The \( x^2+1 \) factor is irreducible over real numbers and requires a \( Cx+D \) numerator in partial fraction decomposition.

Question 16. \( \frac{1}{x\left(x^{n}+1\right)} \)
Answer:To integrate \( \frac{1}{x(x^n+1)} \), we use a substitution method. Multiply the numerator and denominator by \( x^{n-1} \): \( \frac{1}{x(x^n+1)} = \frac{x^{n-1}}{x^n(x^n+1)} \) Now, let \( t = x^n \). Then \( dt = nx^{n-1} dx \), which implies \( x^{n-1} dx = \frac{1}{n} dt \). Substitute these into the integral: \( I = \int \frac{x^{n-1}}{x^n(x^n+1)} dx = \int \frac{1}{t(t+1)} \frac{1}{n} dt = \frac{1}{n} \int \frac{1}{t(t+1)} dt \) Now, we use partial fraction decomposition for \( \frac{1}{t(t+1)} \): \( \frac{1}{t(t+1)} = \frac{A}{t} + \frac{B}{t+1} \) This implies \( 1 = A(t+1) + Bt \) (Equation 2). Put \( t = 0 \) in Equation 2: \( 1 = A(0+1) + B(0) \) So, \( A = 1 \). Put \( t = -1 \) in Equation 2: \( 1 = A(0) + B(-1) \) So, \( B = -1 \). Thus, \( \frac{1}{t(t+1)} = \frac{1}{t} - \frac{1}{t+1} \). Now, integrate with respect to \( t \): \( \int \left( \frac{1}{t} - \frac{1}{t+1} \right) dt = \log|t| - \log|t+1| + C' = \log \left| \frac{t}{t+1} \right| + C' \) Substitute this back into our main integral for \( I \): \( I = \frac{1}{n} \left( \log \left| \frac{t}{t+1} \right| \right) + C \) Finally, substitute back \( t = x^n \): \( I = \frac{1}{n} \log \left| \frac{x^n}{x^n+1} \right| + C \)
In simple words: First, multiply the top and bottom by \( x^{n-1} \). Then, replace \( x^n \) with a new variable, 't'. This changes the integral into a simpler form. Use partial fractions to break it apart, integrate, and then change 't' back to \( x^n \) to get the final answer.

Exam Tip: For integrals of the form \( \int \frac{dx}{x(x^n+1)} \), multiplying the numerator and denominator by \( x^{n-1} \) and then substituting \( t=x^n \) is a common and effective technique.

Question 18. \( \frac{(x^{2}+1)(x^{2}+2)}{(x^{2}+3)(x^{2}+4)} \)
Answer:Let \( x^2 = y \). The expression becomes: \( \frac{(y+1)(y+2)}{(y+3)(y+4)} = \frac{y^2+3y+2}{y^2+7y+12} \) This is an improper fraction (degree of numerator = degree of denominator), so we perform polynomial long division: \[ \frac{y^2+3y+2}{y^2+7y+12} = 1 - \frac{4y+10}{y^2+7y+12} \] Now, we need to decompose \( \frac{4y+10}{y^2+7y+12} \) using partial fractions. Factor the denominator: \( y^2+7y+12 = (y+3)(y+4) \). Let \( \frac{4y+10}{(y+3)(y+4)} = \frac{A}{y+3} + \frac{B}{y+4} \) (Equation 2). This implies \( 4y+10 = A(y+4) + B(y+3) \). Put \( y = -3 \): \( 4(-3)+10 = A(-3+4) + B(0) \) \( -12+10 = A(1) \) So, \( A = -2 \). Put \( y = -4 \): \( 4(-4)+10 = A(0) + B(-4+3) \) \( -16+10 = B(-1) \) \( -6 = -B \) So, \( B = 6 \). Thus, \( \frac{4y+10}{(y+3)(y+4)} = \frac{-2}{y+3} + \frac{6}{y+4} \). Substituting this back into the expression for \( \frac{(y+1)(y+2)}{(y+3)(y+4)} \): \( \frac{(y+1)(y+2)}{(y+3)(y+4)} = 1 - \left( \frac{-2}{y+3} + \frac{6}{y+4} \right) = 1 + \frac{2}{y+3} - \frac{6}{y+4} \) Now, substitute back \( y = x^2 \): \( \frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)} = 1 + \frac{2}{x^2+3} - \frac{6}{x^2+4} \) Now, we integrate both sides: \( \int \frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)} dx = \int 1 dx + \int \frac{2}{x^2+3} dx - \int \frac{6}{x^2+4} dx \) \( = x + 2 \int \frac{1}{x^2+(\sqrt{3})^2} dx - 6 \int \frac{1}{x^2+2^2} dx \) Using the formula \( \int \frac{1}{x^2+a^2} dx = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) \): \( = x + 2 \left( \frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{x}{\sqrt{3}}\right) \right) - 6 \left( \frac{1}{2}\tan^{-1}\left(\frac{x}{2}\right) \right) + C \) \( = x + \frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{x}{\sqrt{3}}\right) - 3\tan^{-1}\left(\frac{x}{2}\right) + C \)
In simple words: First, substitute \( x^2 \) with 'y' to simplify the fraction. Then, divide the top and bottom as it's an improper fraction. Break the remainder into simpler parts using partial fractions. Put \( x^2 \) back in for 'y'. Finally, integrate each part using the standard formulas for inverse tangent.

Exam Tip: The substitution \( x^2=y \) is valid for algebraic manipulation, but remember to substitute back to \( x \) before integrating. Also, correctly identify the 'a' value in the \( \tan^{-1} \) integral formula; for \( x^2+3 \), \( a=\sqrt{3} \), and for \( x^2+4 \), \( a=2 \).

Question 19. \( \frac{2 x}{(x^{2}+1)(x^{2}+3)} \)
Answer:To integrate \( \frac{2x}{(x^2+1)(x^2+3)} \), we use a substitution method. Let \( t = x^2 \). Then \( dt = 2x dx \). Substitute these into the integral: \( I = \int \frac{2x}{(x^2+1)(x^2+3)} dx = \int \frac{1}{(t+1)(t+3)} dt \) Now, we use partial fraction decomposition for \( \frac{1}{(t+1)(t+3)} \): \( \frac{1}{(t+1)(t+3)} = \frac{A}{t+1} + \frac{B}{t+3} \) This implies \( 1 = A(t+3) + B(t+1) \) (Equation 1). Put \( t = -1 \) in Equation 1: \( 1 = A(-1+3) + B(0) \) \( 1 = 2A \) So, \( A = \frac{1}{2} \). Put \( t = -3 \) in Equation 1: \( 1 = A(0) + B(-3+1) \) \( 1 = -2B \) So, \( B = -\frac{1}{2} \). Thus, \( \frac{1}{(t+1)(t+3)} = \frac{\frac{1}{2}}{t+1} - \frac{\frac{1}{2}}{t+3} \). Now, integrate with respect to \( t \): \( \int \left( \frac{1}{2(t+1)} - \frac{1}{2(t+3)} \right) dt = \frac{1}{2} \int \frac{1}{t+1} dt - \frac{1}{2} \int \frac{1}{t+3} dt \) \( = \frac{1}{2}\log|t+1| - \frac{1}{2}\log|t+3| + C' \) \( = \frac{1}{2} \left( \log|t+1| - \log|t+3| \right) + C' \) \( = \frac{1}{2} \log \left| \frac{t+1}{t+3} \right| + C' \) Finally, substitute back \( t = x^2 \): \( I = \frac{1}{2} \log \left| \frac{x^2+1}{x^2+3} \right| + C \)
In simple words: First, substitute \( x^2 \) with 't' and \( 2x dx \) with 'dt' to simplify the integral. Then, break the new fraction into simpler parts. Integrate these parts using the logarithm rule. Finally, change 't' back to \( x^2 \) to get the final answer.

Exam Tip: This type of integral (rational function of \( x^2 \) multiplied by \( 2x dx \)) is a strong indicator for the substitution \( t=x^2 \). It transforms the integral into a standard partial fraction form in terms of \( t \).

Question 20. \( \frac{1}{x\left(x^{4}-1\right)} \)
Answer:To integrate \( \frac{1}{x(x^4-1)} \), we use a substitution method. Multiply the numerator and denominator by \( x^{3} \): \( \frac{1}{x(x^4-1)} = \frac{x^{3}}{x^4(x^4-1)} \) Now, let \( t = x^4 \). Then \( dt = 4x^3 dx \), which implies \( x^3 dx = \frac{1}{4} dt \). Substitute these into the integral: \( I = \int \frac{x^{3}}{x^4(x^4-1)} dx = \int \frac{1}{t(t-1)} \frac{1}{4} dt = \frac{1}{4} \int \frac{1}{t(t-1)} dt \) Now, we use partial fraction decomposition for \( \frac{1}{t(t-1)} \): \( \frac{1}{t(t-1)} = \frac{A}{t} + \frac{B}{t-1} \) This implies \( 1 = A(t-1) + Bt \) (Equation 1). Put \( t = 0 \) in Equation 1: \( 1 = A(0-1) + B(0) \) So, \( A = -1 \). Put \( t = 1 \) in Equation 1: \( 1 = A(0) + B(1) \) So, \( B = 1 \). Thus, \( \frac{1}{t(t-1)} = \frac{-1}{t} + \frac{1}{t-1} \). Now, integrate with respect to \( t \): \( \int \left( \frac{-1}{t} + \frac{1}{t-1} \right) dt = -\log|t| + \log|t-1| + C' = \log \left| \frac{t-1}{t} \right| + C' \) Substitute this back into our main integral for \( I \): \( I = \frac{1}{4} \left( \log \left| \frac{t-1}{t} \right| \right) + C \) Finally, substitute back \( t = x^4 \): \( I = \frac{1}{4} \log \left| \frac{x^4-1}{x^4} \right| + C \)
In simple words: First, multiply the top and bottom of the fraction by \( x^3 \). Then, replace \( x^4 \) with 't' and \( x^3 dx \) with \( \frac{1}{4} dt \). This converts the integral into a simpler form that can be solved using partial fractions. Integrate the simplified expression, and finally, change 't' back to \( x^4 \).

Exam Tip: For integrals of the form \( \int \frac{dx}{x(x^n \pm a)} \), a common strategy is to multiply by \( x^{n-1} \) in the numerator and denominator, then use the substitution \( t = x^n \).

Question 21. Integrate the following function: \( \frac{1}{e^{x}-1} \)
Answer: Let the integral be \( I \). We have: \[ I = \int \frac{dx}{e^{x}-1} \] To make it easier, we multiply the numerator and denominator by \( e^x \): \[ I = \int \frac{e^{x} dx}{e^{x}(e^{x}-1)} \] Now, let's substitute \( t = e^x \). Then, \( dt = e^x dx \). The integral transforms into: \[ I = \int \frac{dt}{t(t-1)} \] We will use partial fractions for \( \frac{1}{t(t-1)} \). Let \[ \frac{1}{t(t-1)} = \frac{A}{t} + \frac{B}{t-1} \] Multiplying by \( t(t-1) \) on both sides gives: \[ 1 = A(t-1) + Bt \] To find A, set \( t = 0 \): \( 1 = A(0-1) + B(0) \) \( 1 = -A \)
\( \implies A = -1 \) To find B, set \( t = 1 \): \( 1 = A(1-1) + B(1) \) \( 1 = B \)
\( \implies B = 1 \) So, the partial fraction decomposition is: \[ \frac{1}{t(t-1)} = \frac{-1}{t} + \frac{1}{t-1} \] Now, substitute this back into the integral: \[ I = \int \left(-\frac{1}{t} + \frac{1}{t-1}\right) dt \] \[ I = -\int \frac{1}{t} dt + \int \frac{1}{t-1} dt \] \[ I = -\log|t| + \log|t-1| + C \] Using logarithm properties, \( \log a - \log b = \log \frac{a}{b} \): \[ I = \log\left|\frac{t-1}{t}\right| + C \] Finally, substitute back \( t = e^x \): \[ I = \log\left|\frac{e^{x}-1}{e^{x}}\right| + C \]In simple words: We changed the variable from x to t by letting \( t = e^x \). Then, we broke the fraction into two simpler ones using partial fractions. After integrating those simple fractions, we put \( e^x \) back in place of t to get the final answer.

Exam Tip: Remember to always substitute back the original variable after integration when using substitution methods. Also, be careful with the signs in partial fraction decomposition.

Question 22. Integrate the following function: \( \int \frac{x}{(x-1)(x-2)} dx \) equals
(A) \( \log \left|\frac{(x-1)^{2}}{x-2}\right| + C \)
(B) \( \log \left|\frac{(x-2)^{2}}{x-1}\right| + C \)
(C) \( \log \left|\frac{(x-1)^{2}}{x-2}\right| + C \)
(D) \( \log |(x - 1)(x - 2)| + C \)
Answer: (B) \( \log \left|\frac{(x-2)^{2}}{x-1}\right| + C \)
Solution: Let's find the partial fractions for the integrand \( \frac{x}{(x-1)(x-2)} \). We can write it as: \[ \frac{x}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2} \] Multiply both sides by \( (x-1)(x-2) \): \[ x = A(x-2) + B(x-1) \] To find A, set \( x = 1 \): \( 1 = A(1-2) + B(1-1) \) \( 1 = A(-1) + 0 \) \( \implies A = -1 \) To find B, set \( x = 2 \): \( 2 = A(2-2) + B(2-1) \) \( 2 = 0 + B(1) \) \( \implies B = 2 \) So, the integrand becomes: \[ \frac{x}{(x-1)(x-2)} = \frac{-1}{x-1} + \frac{2}{x-2} \] Now, we integrate this expression: \[ I = \int \left(\frac{-1}{x-1} + \frac{2}{x-2}\right) dx \] \[ I = -\int \frac{1}{x-1} dx + 2\int \frac{1}{x-2} dx \] \[ I = -\log|x-1| + 2\log|x-2| + C \] Using logarithm properties, \( n \log a = \log a^n \) and \( \log a - \log b = \log \frac{a}{b} \): \[ I = \log(x-2)^2 - \log|x-1| + C \] \[ I = \log\left|\frac{(x-2)^2}{x-1}\right| + C \] This matches option (B).In simple words: First, we split the fraction into two simpler parts. Then, we integrated each part separately. Finally, we used log rules to combine them into one log term.

Exam Tip: Partial fraction decomposition is key for integrating rational functions. Always remember the properties of logarithms to simplify your final answer.

Question 23. Integrate the following function: \( \int \frac{dx}{x\left(x^{2}+1\right)} \) equals
(A) \( \log|x| - \frac{1}{2} \log |x^{2} + 1| + C \)
(B) \( \log|x| + \frac{1}{2} \log|x^{2} + 1| + C \)
(C) \( -\log|x| + \frac{1}{2} \log|x^{2} + 1| + C \)
(D) \( \frac{1}{2} \log|x| + \log|x^{2} + 1| + C \)
Answer: (A) \( \log|x| - \frac{1}{2} \log |x^{2} + 1| + C \)
Solution: Let the integral be \( I \). \[ I = \int \frac{dx}{x(x^{2}+1)} \] To solve this, we can try a substitution. Let \( x^2 = t \). Then, \( 2x dx = dt \), so \( dx = \frac{dt}{2x} \). Since \( x = \sqrt{t} \), we have \( dx = \frac{dt}{2\sqrt{t}} \). Substitute these into the integral: \[ I = \int \frac{1}{\sqrt{t}(t+1)} \frac{dt}{2\sqrt{t}} \] \[ I = \int \frac{1}{2t(t+1)} dt \] \[ I = \frac{1}{2} \int \frac{1}{t(t+1)} dt \] Now, we use partial fractions for \( \frac{1}{t(t+1)} \). Let \[ \frac{1}{t(t+1)} = \frac{A}{t} + \frac{B}{t+1} \] Multiply both sides by \( t(t+1) \): \[ 1 = A(t+1) + Bt \] To find A, set \( t = 0 \): \( 1 = A(0+1) + B(0) \) \( 1 = A \)
\( \implies A = 1 \) To find B, set \( t = -1 \): \( 1 = A(-1+1) + B(-1) \) \( 1 = 0 - B \) \( \implies B = -1 \) So, the partial fraction decomposition is: \[ \frac{1}{t(t+1)} = \frac{1}{t} - \frac{1}{t+1} \] Substitute this back into the integral for \( I \): \[ I = \frac{1}{2} \int \left(\frac{1}{t} - \frac{1}{t+1}\right) dt \] \[ I = \frac{1}{2} \left( \int \frac{1}{t} dt - \int \frac{1}{t+1} dt \right) \] \[ I = \frac{1}{2} \left( \log|t| - \log|t+1| \right) + C \] Using logarithm properties, \( \log a - \log b = \log \frac{a}{b} \): \[ I = \frac{1}{2} \log\left|\frac{t}{t+1}\right| + C \] Finally, substitute back \( t = x^2 \): \[ I = \frac{1}{2} \log\left|\frac{x^{2}}{x^{2}+1}\right| + C \] Using another logarithm property, \( \log a^n = n \log a \): \[ I = \frac{1}{2} (\log|x^2| - \log|x^2+1|) + C \] \[ I = \frac{1}{2} (2\log|x| - \log|x^2+1|) + C \] \[ I = \log|x| - \frac{1}{2} \log|x^2+1| + C \] This matches option (A).In simple words: We used a substitution to simplify the integral, then broke the new fraction into simpler parts. After integrating these parts, we put the original variable back and used log rules to match one of the options.

Exam Tip: For integrals of rational functions involving \( x^2+a^2 \), a substitution like \( t = x^2 \) or partial fractions with \( \frac{Bx+C}{x^2+1} \) is often effective. Ensure you manage logarithmic properties carefully at the end.