#### Provide Solution For  R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise  Revision Exercise Question 122 Maths Textbook Solution.

Answer: $\frac{1}{8} \log \left|\frac{x-1}{x+1}\right|-\left(\frac{3 x-4}{4(x-1)^{2}}\right)+C$

Given:$\int \frac{x^{2}}{(x-1)^{3}(x+1)} d x$

Hint: you must know the steps to integrate by partial function

Explanation: let $I=\int \frac{x^{2}}{(x-1)^{3}(x+1)} d x$

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

Multiplying by $(x-1)^{3}(x+1)$

$x^{2}=A(x-1)^{3}(x+1)+B(x-1)(x+1)+C(x+1)+D(x-1)^{3}$

Putting x=1

$1=A(0)+B(0)+C(2)+D(0) \Rightarrow c=\frac{1}{2}$

Putting x=-1

$1=A(0)+B(0)+C(2)+D(-8) \Rightarrow c=\frac{-1}{8}$

Putting x=0

\begin{aligned} &1=A(1)+B(-1)+C(1)+D(-1)\\ &A-B+C-D=0\\ &A-B+\frac{1}{2}-\left(-\frac{1}{8}\right)=0 \Rightarrow A-B+\frac{1}{2}+\frac{1}{8}=0\\ &A-B+\frac{5}{8}=0 \Rightarrow A-B=-\frac{5}{8} \end{aligned}

Putting x=2

\begin{aligned} &4=A(1)(3)+B(1)(3)+C(3)+D(1) \\ &4=3 A+3 B+3\left(\frac{1}{2}\right)+\left(-\frac{1}{8}\right) \\ &\Rightarrow 3 A+3 B=4-\frac{3}{2}+\frac{1}{8}=>3(A+B)=\frac{32-12+1}{8}=\frac{21}{8} \\ &A+B=\frac{3}{8} \end{aligned}

\begin{aligned} &A+B=\frac{-5}{8}+\frac{7}{8}=\frac{2}{8} \\ &2 A-\frac{2}{8} \Rightarrow A=\frac{1}{8} \end{aligned}

Put in (1)

\begin{aligned} &\frac{1}{8}-B=\frac{5}{8} \Rightarrow B=\frac{1}{8}+\frac{5}{8}=\frac{6}{8}=>B=\frac{3}{4} \\ &\frac{x^{2}}{(x-1)^{3}}=\frac{1}{8(x-1)}+\frac{3}{4(x-1)^{2}}=\frac{1}{2(x-1)^{3}}=\frac{1}{8(x+1)} \\ &\frac{1}{8} \int \frac{d x}{x-1}+\frac{3}{4} \int \frac{(x-1)^{-2+1}}{-2+1}+\frac{1}{2} \frac{(x-1)^{-3+1}}{-3+1}-\frac{1}{8} \log |x+1|+C \\ &=\frac{1}{8} \log \left|\frac{x-1}{x+1}\right|-\frac{3 x-4}{4(x+1)^{2}}+C \end{aligned}

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