NCERT Solutions for Class 12 Maths Chapter 7 Integrals (ex 7.5) Exercise 7.5

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NCERT Solutions for Class 12 Maths Chapter 7 Integrals (Ex 7.5) Exercise 7.5

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Access NCERT Solutions For Class 12 Maths Chapter 7 – Integrals

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NCERT Solutions For Class 12 Maths Chapter 7 Integrals Exercise 7.5

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Q.1

Integrate the rational function \frac{x}{(x + 1) (x + 2)}

Ans

$\begin{array}{l} Let \frac{x}{(x + 1) (x + 2)} = \frac{A}{x + 1} + \frac{B}{x + 2} \\ x = A (x + 2) + B (x + 1) \\ x = (A + B) x + 2 A + B \\ \Rightarrow A + B = 1 and 2 A + B = 0 \\ \Rightarrow A = - 1 and B = 2 \\ So, \frac{x}{(x + 1) (x + 2)} = \frac{- 1}{x + 1} + \frac{2}{x + 2} \\ ∴ \int \frac{x}{(x + 1) (x + 2)} dx = \int \frac{- 1}{x + 1} dx + \int \frac{2}{x + 2} dx \\ = \log {(x + 1)}^{- 1} + \log {(x + 2)}^{2} + C \\ = \log \frac{{(x + 2)}^{2}}{(x + 1)} + C \end{array}$

Q.2

Integrate the rational function \frac{1}{x^{2} - 9}

Ans

$\begin{array}{l} Let \frac{1}{x^{2} - 9} = \frac{1}{(x + 3) (x - 3)} \\ = \frac{A}{x + 3} + \frac{B}{x - 3} \\ 1 = A (x - 3) + B (x + 3) \\ 1 = (A + B) x + 3 (- A + B) \\ \Rightarrow A + B = 0 and 3 (- A + B) = 1 \\ \Rightarrow A = - \frac{1}{6} and B = \frac{1}{6} \\ So, \frac{1}{(x + 3) (x - 3)} = \frac{- 1}{6 (x + 3)} + \frac{1}{6 (x - 3)} \\ ∴ \int \frac{1}{(x + 3) (x - 3)} dx = \int \frac{- 1}{6 (x + 3)} dx + \int \frac{1}{6 (x - 3)} dx \\ = - \frac{1}{6} \log | x + 3 | + \frac{1}{6} \log | x - 3 | + C \\ = \frac{1}{6} \log | \frac{x - 3}{x + 3} | + C \end{array}$

Q.3

Integrate the rational function \frac{3 x - 1}{(x - 1) (x - 2) (x - 3)}

Ans

$\begin{array}{l} Let \\ \frac{3 x - 1}{(x - 1) (x - 2) (x - 3)} \\ = \frac{A}{(x - 1)} + \frac{B}{(x - 2)} + \frac{C}{(x - 3)} \\ 3 x - 1 = A (x - 2) (x - 3) + B (x - 1) (x - 3) + C (x - 1) (x - 2) . .. (i) \\ Substituting x = 1, 2 and 3 respectively in equation (i), we get \\ A = 1, B = - 5 and C = 4 \\ ∴ \frac{3 x - 1}{(x - 1) (x - 2) (x - 3)} = \frac{1}{x - 1} + \frac{- 5}{x - 2} + \frac{4}{(x - 3)} \\ ∴ \int \frac{3 x - 1}{(x - 1) (x - 2) (x - 3)} dx = \int \frac{1}{x - 1} dx - \int \frac{5}{x - 2} dx + \int \frac{4}{(x - 3)} dx \\ = \log |x - 1| - 5 \log |x - 2| + 4 \log |x - 3| + C \end{array}$

Q.4

Integrate the rational function \frac{x}{(x - 1) (x - 2) (x - 3)}

Ans

$\begin{array}{l} Let \frac{x}{(x - 1) (x - 2) (x - 3)} \\ = \frac{A}{(x - 1)} + \frac{B}{(x - 2)} + \frac{C}{(x - 3)} \\ x = A (x - 2) (x - 3) + B (x - 1) (x - 3) + C (x - 1) (x - 2) . .. (i) \\ Substituting x = 1, 2 and 3 respectively in equation (i), we get \\ A = \frac{1}{2}, B = - 2 and C = \frac{3}{2} \\ \Rightarrow \frac{x}{(x - 1) (x - 2) (x - 3)} = \frac{1}{2 (x - 1)} + \frac{- 2}{x - 2} + \frac{3}{2 (x - 3)} \\ ∴ \int \frac{x}{(x - 1) (x - 2) (x - 3)} dx = \frac{1}{2} \int \frac{1}{x - 1} dx - 2 \int \frac{1}{x - 2} dx + \frac{3}{2} \int \frac{1}{(x - 3)} dx \\ = \frac{1}{2} \log | x - 1 | - 2 \log | x - 2 | + \frac{3}{2} \log | x - 3 | + C \end{array}$

Q.5

Integrate the rational function \frac{2 x}{x^{2} + 3 x + 2}

Ans

$\begin{array}{l} Let \frac{2 x}{x^{2} + 3 x + 2} = \frac{2 x}{(x + 1) (x + 2)} \\ = \frac{A}{x + 1} + \frac{B}{x + 2} \\ 2 x = A (x + 2) + B (x + 1) . .. (i) \\ Substituting x = - 1 and - 2 respectively in equation (i), we get \\ A = - 2, B = 4 \\ ∴ \frac{2 x}{(x + 1) (x + 2)} = \frac{- 2}{x + 1} + \frac{4}{x + 2} \\ ∴ \int \frac{2 x}{(x + 1) (x + 2)} dx = \int \frac{- 2}{x + 1} dx + \int \frac{4}{x + 2} dx \\ = - 2 \log | x + 1 | + 4 \log | x + 2 | + C \end{array}$

Q.6

Integrate the rational function \frac{1 - x^{2}}{x (1 - 2 x)}

Ans

$\begin{array}{l} \frac{1 - x^{2}}{x (1 - 2 x)} = \frac{1}{2} + \frac{1}{2} \{\frac{2 - x}{x (1 - 2 x)}\} . .. (i) \\ Let \frac{2 - x}{x (1 - 2 x)} = \frac{A}{x} + \frac{B}{1 - 2 x} \\ \Rightarrow 2 - x = A (1 - 2 x) + Bx \\ Substituting x = 0 and \frac{1}{2} respectively in equation (i), we get \\ A = 2, B = 3 \\ ∴ \frac{2 - x}{x (1 - 2 x)} = \frac{2}{x} + \frac{3}{1 - 2 x} \\ From equation (i), we have \\ \int \frac{1 - x^{2}}{x (1 - 2 x)} dx = \int \frac{1}{2} dx + \frac{1}{2} \int \frac{2}{x} dx + \frac{1}{2} \int \frac{3}{1 - 2 x} dx \\ = \frac{1}{2} x + \frac{1}{2} (2 \log |x| - \frac{3}{4} \log |1 - 2 x|) + C \\ = \frac{1}{2} x + \log |x| - \frac{3}{4} \log |1 - 2 x| + C \end{array}$

Q.7

Integrate the rational function \frac{x}{(x^{2} + 1) (x - 1)}

Ans

$\begin{array}{l} Let \frac{x}{(x^{2} + 1) (x - 1)} = \frac{Ax + B}{x^{2} + 1} + \frac{C}{x - 1} . .. (i) \\ x = (Ax + B) (x - 1) + C (x^{2} + 1) \\ = {Ax}^{2} - Ax + Bx - B + {Cx}^{2} + C \\ = (A + C) x^{2} + (- A + B) x + (- B + C) \\ Equating {coefficients of x}^{2}, x and constant term from both sides, \\ we get \\ A + C = 0, - A + B = 1 and - B + C = 0 \\ On solving these equations, we get \\ A = - \frac{1}{2}, B = \frac{1}{2}, C = \frac{1}{2} \\ So, from equation (i), we get \\ ∴ \frac{x}{(x^{2} + 1) (x - 1)} = \frac{\frac{1}{2} (- x + 1)}{x^{2} + 1} + \frac{\frac{1}{2}}{x - 1} \\ ∴ \int \frac{x}{(x^{2} + 1) (x - 1)} dx = \frac{1}{2} \int \frac{(- x)}{x^{2} + 1} dx + \frac{1}{2} \int \frac{1}{x^{2} + 1} dx + \frac{1}{2} \int \frac{1}{x - 1} dx \\ = - \frac{1}{4} \log | x^{2} + 1 | + \frac{1}{2} \tan^{- 1} x + \frac{1}{2} \log | x - 1 | + C \\ = \frac{1}{2} \log | x - 1 | - \frac{1}{4} \log | x^{2} + 1 | + \frac{1}{2} \tan^{- 1} x + C \end{array}$

Q.8

Integrate the rational function \frac{x}{{(x - 1)}^{2} (x + 2)}

Ans

$\begin{array}{l} Let \frac{x}{{(x - 1)}^{2} (x + 2)} = \frac{A}{(x - 1)} + \frac{B}{{(x - 1)}^{2}} + \frac{C}{(x + 2)} \\ x = A (x - 1) (x + 2) + B (x + 2) + C {(x - 1)}^{2} . .. (i) \\ Substituting x = 1 in equation (i), we get \\ B = \frac{1}{3} \\ Equating {coefficients of x}^{2} and constant term, we get \\ A + C = 0 \\ - 2A + 2 B + C = 0 \\ On solving, we get \\ A = \frac{2}{9} and C = \frac{- 2}{9} \\ \Rightarrow \frac{x}{{(x - 1)}^{2} (x + 2)} = \frac{2}{9 (x - 1)} + \frac{1}{3 {(x - 1)}^{2}} - \frac{2}{9 (x + 2)} \\ ∴ \int \frac{x}{{(x - 1)}^{2} (x + 2)} dx = \frac{2}{9} \int \frac{1}{x - 1} dx + \frac{1}{3} \int \frac{1}{{(x - 1)}^{2}} dx - \frac{2}{9} \int \frac{1}{(x + 2)} dx \\ = \frac{2}{9} \log | x - 1 | + \frac{1}{3} \frac{- 1}{(x - 1)} - \frac{2}{9} \log | x + 2 | + C \\ = \frac{2}{9} \log | \frac{x - 1}{x + 2} | - \frac{1}{3} \frac{1}{(x - 1)} + C \end{array}$

Q.9

Integrate the rational function \frac{3 x + 5}{x^{3} - x^{2} - x + 1}

Ans

$\begin{array}{l} Given, \frac{3 x + 5}{x^{3} - x^{2} - x + 1} = \frac{3 x + 5}{x^{2} (x - 1) - 1 (x - 1)} \\ = \frac{3 x + 5}{(x - 1) (x^{2} - 1)} \\ = \frac{3 x + 5}{(x - 1) (x - 1) (x + 1)} \\ = \frac{3 x + 5}{{(x - 1)}^{2} (x + 1)} \\ Let, \frac{3 x + 5}{{(x - 1)}^{2} (x + 1)} = \frac{A}{(x - 1)} + \frac{B}{{(x - 1)}^{2}} + \frac{C}{(x + 1)} \\ 3 x + 5 = A (x - 1) (x + 1) + B (x + 1) + C {(x - 1)}^{2} \\ = A (x^{2} - 1) + B (x + 1) + C (x^{2} - 2 x + 1) . .. (i) \\ Substituting x = 1 in equation (i), we get \\ B = \frac{8}{2} = 4 \\ Equating {coefficients of x}^{2}, x and constant term,we get \\ A + C = 0 and B - 2 C = 3 \\ On solving these equations, we get \\ A = - \frac{1}{2} and C = \frac{1}{2} \\ ∴ \frac{3 x + 5}{{(x - 1)}^{2} (x + 1)} = \frac{- 1}{2 (x - 1)} + \frac{4}{{(x - 1)}^{2}} + \frac{1}{2 (x + 1)} \\ \Rightarrow \int \frac{3 x + 5}{{(x - 1)}^{2} (x + 1)} dx = - \frac{1}{2} \int \frac{1}{(x - 1)} dx + 4 \int \frac{1}{{(x - 1)}^{2}} dx + \frac{1}{2} \int \frac{1}{(x + 1)} dx \\ = - \frac{1}{2} \log | x - 1 | + 4 (\frac{- 1}{x - 1}) + \frac{1}{2} \log | x + 1 | + C \\ = \frac{1}{2} \log | \frac{x + 1}{x - 1} | - \frac{4}{x - 1} + C \end{array}$

Q.10

Integrate the rational function \frac{2 x - 3}{(x^{2} - 1) (2 x + 3)}

Ans

$\begin{array}{l} \frac{2 x - 3}{(x^{2} - 1) (2 x + 3)} = \frac{2 x - 3}{(x - 1) (x + 1) (2 x + 3)} \\ = \frac{A}{(x + 1)} + \frac{B}{(x - 1)} + \frac{C}{(2 x + 3)} \\ 2 x - 3 = A (x - 1) (2 x + 3) + B (x - 1) (2 x + 3) + C (x - 1) (x + 1) . .. (i) \\ Substituting x = - 1, 1 and - \frac{3}{2} respectively in equation (i), we get \\ A = \frac{5}{2}, B = - \frac{1}{10} and C = - \frac{24}{5} \\ \Rightarrow \frac{2 x - 3}{(x - 1) (x + 1) (2 x + 3)} = \frac{5}{2 (x + 1)} - \frac{1}{10 (x - 1)} - \frac{24}{5 (2 x + 3)} \\ ∴ \int \frac{2 x - 3}{(x^{2} - 1) (2 x + 3)} dx = \frac{5}{2} \int \frac{1}{(x + 1)} dx - \frac{1}{10} \int \frac{1}{(x - 1)} dx \\ - \frac{24}{5} \int \frac{1}{(2 x + 3)} dx \\ = \frac{5}{2} \log | (x + 1) | - \frac{1}{10} \log | (x - 1) | \\ - \frac{24}{5 \times 2} \log | (2 x + 3) | + D \\ = \frac{5}{2} \log | (x + 1) | - \frac{1}{10} \log | (x - 1) | \\ - \frac{12}{5} \log | (2 x + 3) | + D \end{array}$

Q.11

Integrate the rational function \frac{5 x}{(x + 1) (x^{2} - 4)}

Ans

$\begin{array}{l} \frac{5 x}{(x + 1) (x^{2} - 4)} = \frac{5 x}{(x + 1) (x - 2) (x + 2)} \\ = \frac{A}{(x + 1)} + \frac{B}{(x + 2)} + \frac{C}{(x - 2)} \\ 5 x = A (x - 2) (x + 2) + B (x + 1) (x - 2) + C (x + 1) (x + 2) . .. (i) \\ Substituting x = - 1, - 2 and 2 respectively in equation (i), we get \\ A = \frac{5}{3}, B = - \frac{5}{2} and C = \frac{5}{6} \\ \Rightarrow \frac{5 x}{(x + 1) (x - 2) (x + 2)} = \frac{5}{3 (x + 1)} - \frac{5}{2 (x + 2)} + \frac{5}{6 (x - 2)} \\ ∴ \int \frac{2 x - 3}{(x^{2} - 1) (2 x + 3)} dx = \frac{5}{3} \int \frac{1}{(x + 1)} dx - \frac{5}{2} \int \frac{1}{(x + 2)} dx \\ + \frac{5}{6} \int \frac{1}{(x - 2)} dx \\ = \frac{5}{3} \log |(x + 1)| - \frac{5}{2} \log |(x + 2)| \\ + \frac{5}{6} \log |(x - 2)| + D \end{array}$

Q.12

Integrate the rational function \frac{x^{3} + x + 1}{x^{2} - 1}

Ans

$\begin{array}{l} ∵ \frac{x^{3} + x + 1}{x^{2} - 1} = x + \frac{2 x + 1}{x^{2} - 1} \\ Let, \frac{2 x + 1}{x^{2} - 1} = \frac{A}{x + 1} + \frac{B}{x - 1} \\ 2 x + 1 = A (x - 1) + B (x + 1) . .. (i) \\ Substituting x = - 1 and 1 respectively in equation (i), we get \\ 2 (- 1) + 1 = A (- 1 - 1) + B (- 1 + 1) \\ - 1 = A (- 2) \Rightarrow A = \frac{1}{2} \\ and B = \frac{3}{2} \\ \Rightarrow \frac{x^{3} + x + 1}{x^{2} - 1} = x + \frac{1}{2 (x + 1)} + \frac{3}{2 (x - 1)} \\ \int \frac{x^{3} + x + 1}{x^{2} - 1} dx = \int x dx + \frac{1}{2} \int \frac{1}{(x + 1)} dx + \frac{3}{2} \int \frac{1}{x - 1} dx \\ = \frac{x^{2}}{2} + \frac{1}{2} \log | x + 1 | + \frac{3}{2} \log | x - 1 | + C \end{array}$

Q.13

Integrate the rational function \frac{2}{(1 - x) (1 + x^{2})}

Ans

$\begin{array}{l} We have, \frac{2}{(1 - x) (1 + x^{2})} \\ Let \frac{2}{(1 - x) (1 + x^{2})} = \frac{A}{1 - x} + \frac{Bx + C}{1 + x^{2}} \\ 2 = A (1 + x^{2}) + (Bx + C) (1 - x) \\ 2 = x^{2} (A - B) + x (B - C) + (A + C) \\ Equating {the coefficients of x}^{2}, x and constant term, we get \\ A - B = 0 \\ B - C = 0 \\ A + C = 2 \\ On solving these equations, we get \\ A = 1, B = 1 and C = 1 \\ ∴ \frac{2}{(1 - x) (1 + x^{2})} = \frac{1}{1 - x} + \frac{x + 1}{1 + x^{2}} \\ \Rightarrow \int \frac{2}{(1 - x) (1 + x^{2})} dx = \int \frac{1}{1 - x} dx + \int \frac{x + 1}{1 + x^{2}} dx \\ = \int \frac{1}{1 - x} dx + \int \frac{x}{1 + x^{2}} dx + \int \frac{1}{1 + x^{2}} dx \\ = - \log | 1 - x | + \frac{1}{2} \log | 1 + x^{2} | + \tan^{- 1} x + C \end{array}$

Q.14

Integrate the rational function \frac{3 x - 1}{{(x + 2)}^{2}}

Ans

$\begin{array}{l} We have, \frac{3 x - 1}{{(x + 2)}^{2}} \\ Let \frac{3 x - 1}{{(x + 2)}^{2}} = \frac{A}{(x + 2)} + \frac{B}{{(x + 2)}^{2}} \\ 3 x - 1 = A (x + 2) + B \\ Equating {the coefficients of x}^{2}, x and constant term, we get \\ A = 3 \\ 2 A + B = - 1 \\ Solving, the above equations, we get \\ A = 3 and B = - 7 \\ ∴ \frac{3 x - 1}{{(x + 2)}^{2}} = \frac{3}{(x + 2)} - \frac{7}{{(x + 2)}^{2}} \\ \Rightarrow \int \frac{3 x - 1}{{(x + 2)}^{2}} dx = \int \frac{3}{(x + 2)} dx - \int \frac{7}{{(x + 2)}^{2}} dx \\ = 3 \log | x + 2 | - 7 (\frac{- 1}{x + 2}) + C \\ = 3 \log | x + 2 | + \frac{7}{x + 2} + C \end{array}$

Q.15

Integrate the rational function \frac{1}{x^{4} - 1}

Ans

$\begin{array}{l} We have, \frac{1}{x^{4} - 1} = \frac{1}{(x + 1) (x - 1) (x^{2} + 1)} \\ Let \frac{1}{x^{4} - 1} = \frac{A}{(x + 1)} + \frac{B}{(x - 1)} + \frac{Cx + D}{(x^{2} + 1)} \\ 1 = A (x - 1) (x^{2} + 1) + B (x + 1) (x^{2} + 1) + (x^{2} - 1) (Cx + D) \\ 1 = A (x^{3} + x - x^{2} - 1) + B (x^{3} + x + x^{2} + 1) + C (x^{3} - x) + D (x^{2} - 1) \\ 1 = x^{3} (A + B + C) + x^{2} (- A + B + D) + x (A + B - C) + (- A + B - D) \\ Equating {the coefficients of x}^{2}, x and constant term, we get \\ A + B + C = 0 \\ - A + B + D = 0 \\ A + B - C = 0 \\ - A + B - D = 1 \\ Solving, the above equations, we get \\ A = - \frac{1}{4}, B = \frac{1}{4}, C = 0 and D = - \frac{1}{2} \\ ∴ \frac{1}{x^{4} - 1} = \frac{- 1}{4 (x + 1)} + \frac{1}{4 (x - 1)} + \frac{- 1}{2 (x^{2} + 1)} \\ \Rightarrow \int \frac{1}{x^{4} - 1} dx = - \frac{1}{4} \int \frac{1}{x + 1} dx + \frac{1}{4} \int \frac{1}{x - 1} dx - \frac{1}{2} \int \frac{1}{x^{2} + 1} dx \\ = - \frac{1}{4} \log | (x + 1) | + \frac{1}{4} \log | (x - 1) | - \frac{1}{2} \tan^{- 1} x + C \\ = \frac{1}{4} \log | \frac{x - 1}{x + 1} | - \frac{1}{2} \tan^{- 1} x + C \end{array}$

Q.16

Integrate the rational function \frac{1}{x (x^{n} - 1)}

Ans

$\begin{array}{l} We have, \frac{1}{x (x^{n} - 1)} = \frac{1}{x (x^{n} - 1)} \times \frac{x^{n - 1}}{x^{n - 1}} \\ = \frac{x^{n - 1}}{x^{n} (x^{n} - 1)} \\ Let t = x^{n} \Rightarrow \frac{dt}{dx} = {nx}^{n - 1} \Rightarrow \frac{dt}{{nx}^{n - 1}} = dx \\ ∴ \int \frac{1}{x (x^{n} - 1)} dx = \int \frac{x^{n - 1}}{x^{n} (x^{n} - 1)} dx \\ = \frac{1}{n} \int \frac{1}{t (t - 1)} dt \\ Let \frac{1}{t (t - 1)} = \frac{A}{t} + \frac{B}{t - 1} \\ 1 = A (t - 1) + Bt \\ = t (A + B) - A \\ Substituting t = 0 and 1 respectively, we get \\ A = 1 and B = - 1 \\ ∴ \frac{1}{t (t - 1)} = \frac{1}{t} + \frac{- 1}{t - 1} \\ \Rightarrow \int \frac{1}{x (x^{n} - 1)} dx = \frac{1}{n} \int (\frac{1}{t} - \frac{1}{t - 1}) dt \\ = \frac{1}{n} \int \frac{1}{t} dt - \frac{1}{n} \int \frac{1}{t - 1} dt \\ = \frac{1}{n} \log | t | - \frac{1}{n} \log | t - 1 | + C \\ = \frac{1}{n} \log | x^{n} | - \frac{1}{n} \log | x^{n} - 1 | + C \\ = \frac{1}{n} \log | \frac{x^{n}}{x^{n} - 1} | + C \end{array}$

Q.17

Integrate the rational function \frac{cosx}{(1 - sinx) (2 - sinx)}

Ans

$\begin{array}{l} We have, \frac{cosx}{(1 - sinx) (2 - sinx)} \\ Let t = sinx \Rightarrow \frac{dt}{dx} = cosx \Rightarrow \frac{dt}{cosx} = dx \\ ∴ \int \frac{cosx}{(1 - sinx) (2 - sinx)} dx = \int \frac{cosx}{(1 - t) (2 - t)} \frac{dt}{cosx} \\ = \int \frac{1}{(1 - t) (2 - t)} dt \\ Let \frac{1}{(1 - t) (2 - t)} = \frac{A}{1 - t} + \frac{B}{2 - t} \\ 1 = A (2 - t) + B (1 - t) . .. (1) \\ Substituting t = 1 and 2 respectively in eqution (1), we get \\ A = 1 and B = - 1 \\ ∴ \frac{1}{(1 - t) (2 - t)} = \frac{1}{1 - t} - \frac{1}{2 - t} \\ So, \\ \int \frac{cosx}{(1 - sinx) (2 - sinx)} dx = \int \frac{1}{(1 - t) (2 - t)} dt \\ = \int \frac{1}{1 - t} dt - \int \frac{1}{2 - t} dt \\ = - \log | 1 - t | + \log | 2 - t | + C \\ = \log | \frac{2 - t}{1 - t} | + C \\ = \log | \frac{2 - sinx}{1 - sinx} | + C \end{array}$

Q.18

Integrate the rational function \frac{(x^{2} + 1) (x^{2} + 2)}{(x^{2} + 3) (x^{2} + 4)}

Ans

$\begin{array}{l} We have, \\ \frac{(x^{2} + 1) (x^{2} + 2)}{(x^{2} + 3) (x^{2} + 4)} = 1 + \frac{(4 x^{2} + 10)}{(x^{2} + 3) (x^{2} + 4)} \\ Let, \\ \frac{(4 x^{2} + 10)}{(x^{2} + 3) (x^{2} + 4)} = \frac{Ax + B}{(x^{2} + 3)} + \frac{Cx + D}{(x^{2} + 4)} \\ 4 x^{2} + 10 = (Ax + B) (x^{2} + 4) + (Cx + D) (x^{2} + 3) \\ = x^{3} (A + C) + x^{2} (B + D) + x (4 A + 3 C) + (4 B + 3 D) \\ Comparing {the coefficients of x}^{3} {, x}^{2}, x and constant terms, \\ we get \\ A + C = 0 \\ B + D = 4 \\ 4 A + 3 C = 0 \\ 4 B + 3 D = 10 \\ On solving these equations, we get \\ A = 0, B = - 2, C = 0 and D = 6 \\ ∴ \frac{(4 x^{2} + 10)}{(x^{2} + 3) (x^{2} + 4)} = \frac{- 2}{(x^{2} + 3)} + \frac{6}{(x^{2} + 4)} \\ Then, \\ \int \frac{(x^{2} + 1) (x^{2} + 2)}{(x^{2} + 3) (x^{2} + 4)} dx = \int 1 dx - (\int \frac{- 2}{(x^{2} + 3)} dx + 6 \int \frac{1}{(x^{2} + 4)} dx) \\ = x + \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{x}{\sqrt{3}}) - \frac{6}{2} \tan^{- 1} (\frac{x}{2}) + C \\ = x + \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{x}{\sqrt{3}}) - 3 \tan^{- 1} (\frac{x}{2}) + C \end{array}$

Q.19

Integrate the rational function \frac{2 x}{(x^{2} + 1) (x^{2} + 3)}

Ans

$\begin{array}{l} We have, \\ \frac{2 x}{(x^{2} + 1) (x^{2} + 3)} \\ Let t = x^{2} \Rightarrow \frac{dt}{dx} = 2 x \Rightarrow dx = \frac{dt}{2 x} \\ \int \frac{2 x}{(x^{2} + 1) (x^{2} + 3)} dx = \int \frac{2 x}{(t + 1) (t + 3)} \frac{dt}{2 x} \\ = \int \frac{1}{(t + 1) (t + 3)} dt \\ Let \frac{1}{(t + 1) (t + 3)} = \frac{A}{t + 1} + \frac{B}{t + 3} \\ 1 = A (t + 3) + B (t + 1) \\ Substituting t = - 1 and - 3 respectively, we get \\ A = \frac{1}{2} and B = - \frac{1}{2} \\ ∴ \frac{1}{(t + 1) (t + 3)} = \frac{1}{2 (t + 1)} - \frac{1}{2 (t + 3)} \\ \Rightarrow \int \frac{1}{(t + 1) (t + 3)} dx = \int \frac{1}{2 (t + 1)} dt - \int \frac{1}{2 (t + 3)} dx \\ = \frac{1}{2} \log | t + 1 | - \frac{1}{2} \log | t + 3 | + C \\ = \frac{1}{2} \log | \frac{t + 1}{t + 3} | + C \\ = \frac{1}{2} \log | \frac{x^{2} + 1}{x^{2} + 3} | + C \end{array}$

Q.20

Integrate the rational function \frac{1}{x (x^{4} - 1)}

Ans

$\begin{array}{l} We have, \frac{1}{x (x^{4} - 1)} = \frac{1}{x (x^{4} - 1)} \times \frac{x^{3}}{x^{3}} \\ = \frac{x^{3}}{x^{4} (x^{4} - 1)} \\ Let t = x^{4} \Rightarrow \frac{dt}{dx} = 4 x^{4 - 1} \Rightarrow \frac{dt}{4 x^{3}} = dx \\ ∴ \int \frac{1}{x (x^{4} - 1)} dx = \int \frac{x^{3}}{x^{4} (x^{4} - 1)} dx \\ = \int \frac{x^{3}}{t (t - 1)} \frac{dt}{4 x^{3}} \\ = \frac{1}{4} \int \frac{1}{t (t - 1)} dt \\ Let \frac{1}{t (t - 1)} = \frac{A}{t} + \frac{B}{t - 1} \\ 1 = A (t - 1) + Bt \\ Substituting t = 0 and 1 respectively, we get \\ A = - 1 and B = 1 \\ ∴ \frac{1}{t (t - 1)} = \frac{- 1}{t} + \frac{1}{t - 1} \\ \Rightarrow \int \frac{1}{x (x^{4} - 1)} dx = \frac{1}{4} \int (- \frac{1}{t} + \frac{1}{t - 1}) dt \\ = - \frac{1}{4} \int \frac{1}{t} dt + \frac{1}{4} \int \frac{1}{t - 1} dt \\ = - \frac{1}{4} \log | t | + \frac{1}{4} \log | t - 1 | + C \\ = - \frac{1}{4} \log | x^{4} | + \frac{1}{4} \log | x^{4} - 1 | + C \\ = \frac{1}{4} \log | \frac{x^{4} - 1}{x^{4}} | + C \end{array}$

Q.21

Integrate the rational function \frac{1}{(e^{x} - 1)}

Ans

$\begin{array}{l} \int \frac{1}{(e^{x} - 1)} dx \\ Let t = e^{x} \Rightarrow \frac{dt}{dx} = e^{x} \\ dx = \frac{dt}{e^{x}} = \frac{dt}{t} \\ ∴ \int \frac{1}{(e^{x} - 1)} dx = \int \frac{1}{(t - 1)} \frac{dt}{t} \\ = \int \frac{1}{t (t - 1)} dt \\ Let \frac{1}{t (t - 1)} = \frac{A}{t} + \frac{B}{t - 1} \\ = \frac{A (t - 1) + Bt}{t (t - 1)} \\ \Rightarrow 1 = A (t - 1) + Bt . .. (i) \\ Putting t = 0 and 1 respectively in equation (i), we get \\ A = - 1 and B = 1 \\ ∴ \frac{1}{t (t - 1)} = \frac{- 1}{t} + \frac{1}{t - 1} \\ \Rightarrow \int \frac{1}{t (t - 1)} dt = \int \frac{- 1}{t} dt + \int \frac{1}{t - 1} dt \\ = - \log | t | + \log | t - 1 | + C \\ = \log | \frac{t - 1}{t} | + C \\ ∴ \int \frac{1}{(e^{x} - 1)} dx = \log | \frac{e^{x} - 1}{e^{x}} | + C \end{array}$

Q.22

$\begin{array}{l} Choose the correct answer \\ \int \frac{xdx}{(x - 1) (x - 2)} dx equals \\ (A) \log |\frac{{(x - 1)}^{2}}{x - 2}| + C \\ (B) \log |\frac{{(x - 2)}^{2}}{x - 1}| + C \\ (C) \log |{(\frac{x - 1}{x - 2})}^{2}| + C \\ (D) \log |(x - 1) (x - 2)| + C \end{array}$

Ans

$\begin{array}{l} We have, \int \frac{xdx}{(x - 1) (x - 2)} dx \\ Let \frac{x}{(x - 1) (x - 2)} = \frac{A}{x - 1} + \frac{B}{x - 2} \\ = \frac{A (x - 2) + B (x - 1)}{(x - 1) (x - 2)} \\ \Rightarrow x = A (x - 2) + B (x - 1) . .. (i) \\ Substituting x = 1 and 2, in equation (i), we get \\ A = - 1 and B = 2 \\ \Rightarrow \frac{x}{(x - 1) (x - 2)} = \frac{- 1}{x - 1} + \frac{2}{x - 2} \\ ∴ \int \frac{x}{(x - 1) (x - 2)} dx = \int \frac{- 1}{x - 1} dx + \int \frac{2}{x - 2} dx \\ = - log | x - 1 | + 2 \log | x - 2 | + C \\ = \log | \frac{{(x - 2)}^{2}}{(x - 1)} | + C \\ Hence, the option (B) is correct. \end{array}$

Q.23

$\begin{array}{l} Choose the correct answer \\ \int \frac{dx}{x (x^{2} + 1)} equals \\ (A) \log |x| - \frac{1}{2} \log (x^{2} + 1) + C \\ (B) \log |x| + \frac{1}{2} \log (x^{2} + 1) + C \\ (C) - \log |x| + \frac{1}{2} \log (x^{2} + 1) + C \\ (D) \frac{1}{2} \log |x| + \log (x^{2} + 1) + C \end{array}$

Ans

$\begin{array}{l} We have, \int \frac{dx}{x (x^{2} + 1)} \\ Let \frac{1}{x (x^{2} + 1)} = \frac{A}{x} + \frac{Bx + C}{(x^{2} + 1)} \\ = \frac{A (x^{2} + 1) + (Bx + C) x}{x (x^{2} + 1)} \\ \Rightarrow 1 = A (x^{2} + 1) + ({Bx}^{2} + Cx) \\ \Rightarrow 1 = x^{2} (A + B) + Cx + A \\ Equating {the coefficients of x}^{2}, x and constant term, we get \\ A + B = 0, C = 0 and A = 1 \\ On solving these equations, we get \\ A = 1, B = - 1 and C = 0 \\ \Rightarrow \frac{1}{x (x^{2} + 1)} = \frac{1}{x} + \frac{- x}{(x^{2} + 1)} \\ ∴ \int \frac{1}{x (x^{2} + 1)} dx = \int \frac{1}{x} dx - \int \frac{x}{(x^{2} + 1)} dx \\ = log | x | - 2 \log | x^{2} + 1 | + C \\ Thus, the correct option is (A) . \end{array}$

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NCERT Solutions for Class 12 Maths Chapter 7 Integrals (ex 7.5) Exercise 7.5