# Evaluate the following integral:

Question:

Evaluate the following integral:

$\int \frac{1}{x\left(x^{4}-1\right)} d x$

Solution:

Let, $\mathrm{I}=\int \frac{1}{\mathrm{x}\left(\mathrm{x}^{4}-1\right)} \mathrm{dx}$

Let, $\frac{1}{x\left(x^{4}-1\right)}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{x-1}+\frac{D}{x^{2}+1}$

$\Rightarrow 1=A(x+1)(x-1)\left(x^{2}+1\right)+B x(x-1)\left(x^{2}+1\right)+c x(x+1)\left(x^{2}+1\right)+D x(x+1)(x-1)$

For, $x=0, A=-1$

For, $x=1, C=\frac{1}{4}$

$\therefore \mathrm{I}=-\int \frac{\mathrm{dx}}{\mathrm{x}}+\frac{1}{4} \int \frac{\mathrm{dx}}{\mathrm{x}+1}+\frac{1}{4} \int \frac{\mathrm{dx}}{\mathrm{x}-1}+\frac{1}{4} \int \frac{\mathrm{dx}}{\mathrm{x}^{2}+1}$

$\Rightarrow \mathrm{I}=-\ln |\mathrm{x}|+\frac{1}{4} \ln |(\mathrm{x}+1)|+\frac{1}{4} \ln |\mathrm{x}-1|+\frac{1}{4} \tan ^{-1} \mathrm{x}+\mathrm{c}$

$\Rightarrow \mathrm{I}=-\ln |\mathrm{x}|+\frac{1}{4}\left(\ln \left|\mathrm{x}^{2}-1\right|\right)+\frac{1}{4} \tan ^{-1} \mathrm{x}+\mathrm{c}$

$\Rightarrow \mathrm{I}=-\frac{1}{4} \ln \left|\mathrm{x}^{4}\right|+\frac{1}{4} \ln \left(\mathrm{x}^{2}-1\right)+\frac{1}{4} \tan ^{-1} \mathrm{x}+\mathrm{c}$

$\Rightarrow \mathrm{I}=\frac{1}{4} \ln \left|\frac{\mathrm{x}^{2}-1}{\mathrm{x}^{4}}\right|++\frac{1}{4} \tan ^{-1} \mathrm{x}+\mathrm{c}$

$\Rightarrow \mathrm{I}=\frac{1}{4} \ln \left|\frac{\mathrm{x}^{2}-1}{\mathrm{x}^{4}}\right|++\frac{1}{4} \tan ^{-1} \mathrm{x}+\mathrm{c}$

Thus, $\int \frac{1}{x\left(x^{4}-1\right)} d x=\frac{1}{4} \ln \left|\frac{x^{4}-1}{x^{4}}\right|+c$