#### explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 55

$\frac{1}{4} \log \left|\frac{x-1}{x+1}\right|-\frac{1}{2} \tan ^{-1} x+C$

Hint:

To solve this integration, we use partial fraction method

Given:

$\int \frac{1}{x^{4}-1} d x \\$

Explanation:

Let

\begin{aligned} &I=\int \frac{1}{x^{4}-1} d x \\ &I=\int \frac{1}{(x-1)(x+1)\left(x^{2}+1\right)} d x \\ &\frac{1}{(x-1)(x+1)\left(x^{2}+1\right)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C x+D}{x^{2}+1} \\ &1=A\left(x^{2}+1\right)(x+1)+B(x-1)\left(x^{2}+1\right)+(C x+D)\left(x^{2}-1\right) \end{aligned}

Put $x=-1$

\begin{aligned} &1=A(0)+B(-2)(2)+(C x+D)(0) \\ &1=-4 B \\ &B=\frac{-1}{4} \end{aligned}

Put $x=1$

\begin{aligned} &1=A(2)(2)+B(0)+(C+D)(0) \\ &1=4 A \\ &A=\frac{1}{4} \end{aligned}

Put $x=0$

\begin{aligned} &1=A-B+D(-1) \\ &1=\frac{1}{4}+\frac{1}{4}-D \\ &1=\frac{1}{2}-D \\ &D=\frac{-1}{2} \end{aligned}

Put $x=2$

\begin{aligned} &1=A(5)(3)+B(1)(5)+(5 C+D)(3) \\ &1=15 A+5 B+15 C+3 D \\ &1=\frac{15}{4}-\frac{5}{4}+15 C-\frac{3}{2} \\ &1=\frac{10}{4}-\frac{3}{2}+15 C \\ &1=\frac{10-6}{4}+15 C \end{aligned}

\begin{aligned} &1=\frac{4}{4}+15 C \\ &1=1+15 C \\ &C=0 \\ &\frac{1}{(x-1)(x+1)\left(x^{2}+1\right)}=\frac{1}{4(x-1)}-\frac{1}{4(x+1)}-\frac{1}{2\left(x^{2}+1\right)} \end{aligned}

Thus

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