#### Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.30 Question 31

$\frac{1}{4} \log |x-1|+\frac{3}{4} \log |x+1|+\frac{1}{2(x+1)}+C$

Hint:

To solve this integration, we use partial fraction method

Given:

$\int \frac{x^{2}}{(x-1)(x+1)^{2}} d x \\$

Explanation:

Let

\begin{aligned} &I=\int \frac{x^{2}}{(x-1)(x+1)^{2}} d x \\ &\frac{x^{2}}{(x-1)(x+1)^{2}}=\frac{A}{x+1}+\frac{B}{x-1}+\frac{C}{(x+1)^{2}} \end{aligned}

\begin{aligned} &x^{2}=A(x+1)(x-1)+B(x+1)^{2}+C(x-1) \\ &x^{2}=x^{2}(A+B)+x(2 B+C)+(-A+B-C) \end{aligned}

Equating similar terms

$A+B=1$                           (1)

$2B+C=0$                       (2)

\begin{aligned} &C=-2 B \\ &-A+B-C=0 \end{aligned}                (3)

\begin{aligned} &-A+B+2 B=0 \\ &-A+3 B=0 \end{aligned}                      (4)

$\begin{gathered} 4 B=1 \\ B=\frac{1}{4} \end{gathered}$

Equation (2)

\begin{aligned} &C=-2 \times \frac{1}{4} \\ &C=\frac{-1}{2} \end{aligned}

Equation (1)

\begin{aligned} &A=1-\frac{1}{4} \\ &A=\frac{3}{4} \end{aligned}

\begin{aligned} &\frac{x^{2}}{(x-1)(x+1)^{2}}=\frac{3}{4(x+1)}+\frac{1}{4(x-1)}-\frac{1}{2(x+1)^{2}} \\ &I=\frac{3}{4} \int \frac{1}{x+1} d x+\frac{1}{4} \int \frac{1}{x-1} d x-\frac{1}{2} \int \frac{1}{(x+1)^{2}} d x \\ &I=\frac{3}{4} \log |x+1|+\frac{1}{4} \log |x-1|+\frac{1}{2(x+1)}+C \end{aligned}