Class 12-NCERT Solutions-Chapter-07-Integrals-Ex 7.5

Solution:

Let \( \frac{x}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2} \).

\[ x = A(x+2) + B(x+1) \]

Put \( x = -1 \): \( -1 = A(1) + 0 \Rightarrow A = -1 \).

Put \( x = -2 \): \( -2 = 0 + B(-1) \Rightarrow B = 2 \).

\[ \begin{aligned} \int \frac{x}{(x+1)(x+2)} \, dx &= \int \left( \frac{-1}{x+1} + \frac{2}{x+2} \right) \, dx \\ &= -\log|x+1| + 2\log|x+2| + C \\ &= \log(x+2)^2 - \log|x+1| + C \\ &= \log \left| \frac{(x+2)^2}{x+1} \right| + C \end{aligned} \]

Solution:

Factor the denominator: \( x^2 - 9 = (x-3)(x+3) \).

Let \( \frac{1}{(x-3)(x+3)} = \frac{A}{x-3} + \frac{B}{x+3} \).

\( 1 = A(x+3) + B(x-3) \).

Put \( x = 3 \): \( 1 = 6A \Rightarrow A = \frac{1}{6} \).

Put \( x = -3 \): \( 1 = -6B \Rightarrow B = -\frac{1}{6} \).

\[ \begin{aligned} \int \frac{1}{x^2-9} \, dx &= \frac{1}{6} \int \frac{dx}{x-3} - \frac{1}{6} \int \frac{dx}{x+3} \\ &= \frac{1}{6} \log|x-3| - \frac{1}{6} \log|x+3| + C \\ &= \frac{1}{6} \log \left| \frac{x-3}{x+3} \right| + C \end{aligned} \]

Solution:

Let \( \frac{3x-1}{(x-1)(x-2)(x-3)} = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3} \).

\( 3x-1 = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2) \).

Put \( x = 1 \): \( 2 = A(-1)(-2) \Rightarrow 2 = 2A \Rightarrow A = 1 \).

Put \( x = 2 \): \( 5 = B(1)(-1) \Rightarrow 5 = -B \Rightarrow B = -5 \).

Put \( x = 3 \): \( 8 = C(2)(1) \Rightarrow 8 = 2C \Rightarrow C = 4 \).

\[ \begin{aligned} \int \frac{3x-1}{(x-1)(x-2)(x-3)} \, dx &= \int \frac{1}{x-1} dx - 5 \int \frac{1}{x-2} dx + 4 \int \frac{1}{x-3} dx \\ &= \log|x-1| - 5\log|x-2| + 4\log|x-3| + C \end{aligned} \]

Solution:

Let \( \frac{x}{(x-1)(x-2)(x-3)} = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3} \).

\( x = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2) \).

Put \( x = 1 \): \( 1 = 2A \Rightarrow A = \frac{1}{2} \).

Put \( x = 2 \): \( 2 = -B \Rightarrow B = -2 \).

Put \( x = 3 \): \( 3 = 2C \Rightarrow C = \frac{3}{2} \).

\[ \begin{aligned} I &= \frac{1}{2}\log|x-1| - 2\log|x-2| + \frac{3}{2}\log|x-3| + C \end{aligned} \]

Solution:

Denominator factors as \( (x+1)(x+2) \).

Let \( \frac{2x}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2} \).

\( 2x = A(x+2) + B(x+1) \).

Put \( x = -1 \): \( -2 = A(1) \Rightarrow A = -2 \).

Put \( x = -2 \): \( -4 = B(-1) \Rightarrow B = 4 \).

\[ \begin{aligned} \int \frac{2x}{x^2+3x+2} \, dx &= \int \frac{-2}{x+1} \, dx + \int \frac{4}{x+2} \, dx \\ &= -2\log|x+1| + 4\log|x+2| + C \end{aligned} \]

Solution:

Rewrite integrand: \( \frac{1-x^2}{x-2x^2} = \frac{x^2-1}{2x^2-x} \). This is an improper fraction (degree of numerator = degree of denominator).

Divide \( x^2-1 \) by \( 2x^2-x \): Quotient is \( \frac{1}{2} \), Remainder is \( \frac{x}{2} - 1 \).

\[ \frac{1-x^2}{x(1-2x)} = \frac{1}{2} + \frac{\frac{1}{2}x-1}{x(1-2x)} = \frac{1}{2} + \frac{x-2}{2x(1-2x)} \]

Let \( \frac{x-2}{x(1-2x)} = \frac{A}{x} + \frac{B}{1-2x} \).

\( x-2 = A(1-2x) + Bx \).

Put \( x = 0 \): \( -2 = A \).

Put \( x = \frac{1}{2} \): \( \frac{1}{2}-2 = \frac{B}{2} \Rightarrow -\frac{3}{2} = \frac{B}{2} \Rightarrow B = -3 \).

\[ \begin{aligned} I &= \int \frac{1}{2} \, dx + \frac{1}{2} \int \left( \frac{-2}{x} - \frac{3}{1-2x} \right) \, dx \\ &= \frac{x}{2} - \log|x| - \frac{3}{2} \frac{\log|1-2x|}{-2} + C \\ &= \frac{x}{2} - \log|x| + \frac{3}{4}\log|1-2x| + C \end{aligned} \]

Solution:

Let \( \frac{x}{(x-1)(x^2+1)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+1} \).

\( x = A(x^2+1) + (Bx+C)(x-1) \).

Put \( x = 1 \): \( 1 = 2A \Rightarrow A = \frac{1}{2} \).

Equating coeffs of \( x^2 \): \( A + B = 0 \Rightarrow B = -A = -\frac{1}{2} \).

Equating constant terms: \( A - C = 0 \Rightarrow C = A = \frac{1}{2} \).

\[ \begin{aligned} I &= \int \left( \frac{1}{2(x-1)} + \frac{-\frac{1}{2}x + \frac{1}{2}}{x^2+1} \right) \, dx \\ &= \frac{1}{2}\log|x-1| - \frac{1}{2} \int \frac{x}{x^2+1} \, dx + \frac{1}{2} \int \frac{1}{x^2+1} \, dx \\ &= \frac{1}{2}\log|x-1| - \frac{1}{4}\log(x^2+1) + \frac{1}{2}\tan^{-1}x + C \end{aligned} \]

Solution:

Let \( \frac{x}{(x-1)^2(x+2)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+2} \).

\( x = A(x-1)(x+2) + B(x+2) + C(x-1)^2 \).

Put \( x = 1 \): \( 1 = 3B \Rightarrow B = \frac{1}{3} \).

Put \( x = -2 \): \( -2 = 9C \Rightarrow C = -\frac{2}{9} \).

Equating coeff of \( x^2 \): \( A + C = 0 \Rightarrow A = -C = \frac{2}{9} \).

\[ \begin{aligned} I &= \frac{2}{9}\log|x-1| + \frac{1}{3} \int (x-1)^{-2} \, dx - \frac{2}{9}\log|x+2| + C \\ &= \frac{2}{9}\log|x-1| - \frac{1}{3(x-1)} - \frac{2}{9}\log|x+2| + C \end{aligned} \]

Solution:

Denominator: \( x^2(x-1) - 1(x-1) = (x^2-1)(x-1) = (x+1)(x-1)^2 \).

Let \( \frac{3x+5}{(x+1)(x-1)^2} = \frac{A}{x+1} + \frac{B}{x-1} + \frac{C}{(x-1)^2} \).

\( 3x+5 = A(x-1)^2 + B(x-1)(x+1) + C(x+1) \).

Put \( x = 1 \): \( 8 = 2C \Rightarrow C = 4 \).

Put \( x = -1 \): \( 2 = 4A \Rightarrow A = \frac{1}{2} \).

Equating \( x^2 \) coeff: \( A + B = 0 \Rightarrow B = -A = -\frac{1}{2} \).

\[ \begin{aligned} I &= \frac{1}{2}\log|x+1| - \frac{1}{2}\log|x-1| - \frac{4}{x-1} + C \end{aligned} \]

Solution:

\( \frac{2x-3}{(x-1)(x+1)(2x+3)} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{C}{2x+3} \).

\( 2x-3 = A(x+1)(2x+3) + B(x-1)(2x+3) + C(x^2-1) \).

Put \( x = 1 \): \( -1 = A(2)(5) \Rightarrow A = -\frac{1}{10} \).

Put \( x = -1 \): \( -5 = B(-2)(1) \Rightarrow B = \frac{5}{2} \).

Put \( x = -\frac{3}{2} \): \( -6 = C(\frac{5}{4}) \Rightarrow C = -\frac{24}{5} \).

\[ \begin{aligned} I &= -\frac{1}{10}\log|x-1| + \frac{5}{2}\log|x+1| - \frac{24}{5} \frac{\log|2x+3|}{2} + C \\ &= -\frac{1}{10}\log|x-1| + \frac{5}{2}\log|x+1| - \frac{12}{5}\log|2x+3| + C \end{aligned} \]

Solution:

\( \frac{5x}{(x+1)(x-2)(x+2)} = \frac{A}{x+1} + \frac{B}{x-2} + \frac{C}{x+2} \).

\( 5x = A(x^2-4) + B(x+1)(x+2) + C(x+1)(x-2) \).

Put \( x = -1 \): \( -5 = -3A \Rightarrow A = \frac{5}{3} \).

Put \( x = 2 \): \( 10 = 12B \Rightarrow B = \frac{5}{6} \).

Put \( x = -2 \): \( -10 = 4C \Rightarrow C = -\frac{5}{2} \).

\[ I = \frac{5}{3}\log|x+1| + \frac{5}{6}\log|x-2| - \frac{5}{2}\log|x+2| + C \]

Solution:

Improper fraction. Divide \( x^3+x+1 \) by \( x^2-1 \): Quotient \( x \), Remainder \( 2x+1 \).

\( \frac{x^3+x+1}{x^2-1} = x + \frac{2x+1}{x^2-1} = x + \frac{2x}{x^2-1} + \frac{1}{x^2-1} \).

\[ \begin{aligned} I &= \int x \, dx + \int \frac{2x}{x^2-1} \, dx + \int \frac{1}{x^2-1} \, dx \\ &= \frac{x^2}{2} + \log|x^2-1| + \frac{1}{2}\log\left|\frac{x-1}{x+1}\right| + C \end{aligned} \]

Solution:

Let \( \frac{2}{(1-x)(1+x^2)} = \frac{A}{1-x} + \frac{Bx+C}{1+x^2} \).

\( 2 = A(1+x^2) + (Bx+C)(1-x) \).

Put \( x = 1 \): \( 2 = 2A \Rightarrow A = 1 \).

Const term: \( 2 = A + C \Rightarrow C = 1 \).

Coeff \( x^2 \): \( A - B = 0 \Rightarrow B = 1 \).

\[ \begin{aligned} I &= \int \frac{1}{1-x} \, dx + \int \frac{x}{1+x^2} \, dx + \int \frac{1}{1+x^2} \, dx \\ &= -\log|1-x| + \frac{1}{2}\log(1+x^2) + \tan^{-1}x + C \end{aligned} \]

Solution:

Let \( \frac{3x-1}{(x+2)^2} = \frac{A}{x+2} + \frac{B}{(x+2)^2} \).

\( 3x-1 = A(x+2) + B \).

Put \( x = -2 \): \( -7 = B \).

Coeff \( x \): \( 3 = A \).

\[ \begin{aligned} I &= 3\log|x+2| + \int \frac{-7}{(x+2)^2} \, dx \\ &= 3\log|x+2| + \frac{7}{x+2} + C \end{aligned} \]

Solution:

\( x^4-1 = (x^2-1)(x^2+1) = (x-1)(x+1)(x^2+1) \).

\( \frac{1}{x^4-1} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{Cx+D}{x^2+1} \).

Solving yields: \( A = \frac{1}{4}, B = -\frac{1}{4}, C = 0, D = -\frac{1}{2} \).

\[ \begin{aligned} I &= \frac{1}{4}\log|x-1| - \frac{1}{4}\log|x+1| - \frac{1}{2}\tan^{-1}x + C \\ &= \frac{1}{4}\log\left|\frac{x-1}{x+1}\right| - \frac{1}{2}\tan^{-1}x + C \end{aligned} \]

Solution:

Multiply numerator and denominator by \( x^{n-1} \): \( \frac{x^{n-1}}{x^n(x^n+1)} \).

Let \( x^n = t \Rightarrow nx^{n-1}dx = dt \).

\[ \begin{aligned} I &= \frac{1}{n} \int \frac{dt}{t(t+1)} \\ &= \frac{1}{n} \int \left( \frac{1}{t} - \frac{1}{t+1} \right) dt \\ &= \frac{1}{n} (\log|t| - \log|t+1|) + C \\ &= \frac{1}{n} \log \left| \frac{x^n}{x^n+1} \right| + C \end{aligned} \]

Solution:

Let \( \sin x = t \Rightarrow \cos x \, dx = dt \).

\[ I = \int \frac{dt}{(1-t)(2-t)} \]

Partial fractions: \( \frac{1}{(1-t)(2-t)} = \frac{1}{1-t} - \frac{1}{2-t} \).

\[ \begin{aligned} I &= \int \frac{dt}{1-t} - \int \frac{dt}{2-t} \\ &= -\log|1-t| - (-\log|2-t|) + C \\ &= \log|2-t| - \log|1-t| + C \\ &= \log \left| \frac{2-\sin x}{1-\sin x} \right| + C \end{aligned} \]

Solution:

Let \( x^2 = y \) for partial fraction decomposition (not integration). \( \frac{(y+1)(y+2)}{(y+3)(y+4)} = \frac{y^2+3y+2}{y^2+7y+12} \).

Divide: \( 1 + \frac{-4y-10}{(y+3)(y+4)} = 1 - \frac{4y+10}{(y+3)(y+4)} \).

\( \frac{4y+10}{(y+3)(y+4)} = \frac{A}{y+3} + \frac{B}{y+4} \).

\( A = \frac{-2}{1} = -2 \). \( B = \frac{-6}{-1} = 6 \).

So integrand is \( 1 - \left( \frac{-2}{x^2+3} + \frac{6}{x^2+4} \right) = 1 + \frac{2}{x^2+3} - \frac{6}{x^2+4} \).

\[ \begin{aligned} I &= x + \frac{2}{\sqrt{3}}\tan^{-1}\frac{x}{\sqrt{3}} - \frac{6}{2}\tan^{-1}\frac{x}{2} + C \\ &= x + \frac{2}{\sqrt{3}}\tan^{-1}\frac{x}{\sqrt{3}} - 3\tan^{-1}\frac{x}{2} + C \end{aligned} \]

Solution:

Let \( x^2 = t \Rightarrow 2x \, dx = dt \).

\[ I = \int \frac{dt}{(t+1)(t+3)} \]

Partial fractions: \( \frac{1}{(t+1)(t+3)} = \frac{1}{2} \left( \frac{1}{t+1} - \frac{1}{t+3} \right) \).

\[ \begin{aligned} I &= \frac{1}{2} (\log|t+1| - \log|t+3|) + C \\ &= \frac{1}{2} \log \left( \frac{x^2+1}{x^2+3} \right) + C \end{aligned} \]

Solution:

Multiply num/den by \( x^3 \): \( \frac{x^3}{x^4(x^4-1)} \). Let \( x^4=t \).

\[ \begin{aligned} I &= \frac{1}{4} \int \frac{dt}{t(t-1)} \\ &= \frac{1}{4} \int \left( \frac{1}{t-1} - \frac{1}{t} \right) dt \\ &= \frac{1}{4} \log \left| \frac{x^4-1}{x^4} \right| + C \end{aligned} \]

Solution:

Let \( e^x = t \Rightarrow e^x dx = dt \Rightarrow dx = \frac{dt}{t} \).

\[ \begin{aligned} I &= \int \frac{dt}{t(t-1)} \\ &= \int \left( \frac{1}{t-1} - \frac{1}{t} \right) dt \\ &= \log|t-1| - \log|t| + C \\ &= \log \left| \frac{e^x-1}{e^x} \right| + C = \log|1-e^{-x}| + C \end{aligned} \]

Answer: (B)

Solution:

\( \frac{x}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2} \).

A (at x=1) = \( 1/-1 = -1 \). B (at x=2) = \( 2/1 = 2 \).

\( I = -\log|x-1| + 2\log|x-2| = \log \frac{(x-2)^2}{|x-1|} \).

Answer: (A)

Solution:

\( \frac{1}{x(x^2+1)} = \frac{A}{x} + \frac{Bx+C}{x^2+1} \). A=1, B=-1, C=0.

\( \int (\frac{1}{x} - \frac{x}{x^2+1}) dx = \log|x| - \frac{1}{2}\log(x^2+1) + C \).