Question

Solve : $\int \frac{x^{4}dx}{(1+x^2{)}^3}$ ?

Answer 1

Given the integral,

$\int \frac{x^{4}dx}{{(x^2+1)}^3}$

using partial fraction we get,

$=\int (\frac{1}{x^2+1}-\frac{2}{{(x^2+1)}^2}+\frac{1}{{(x^2+1)}^3})=\int \frac{1}{x^2+1}dx-2\int \frac{1}{{(x^2+1)}^2}dx+\int \frac{1}{{(x^2+1)}^3}dx$

Here,

$\int \frac{1}{x^2+1}dx={\tan }^{-1}\left(x\right)$

For,

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

applying reduction formula we get,

$=\frac{x}{2(x^2+1)}+\frac{1}{2}\int \frac{1}{x^2+1}dx=\frac{{\tan }^{-1}\left(x\right)}{2}+\frac{x}{2(x^2+1)}$

Again for,

$\int \frac{1}{{(x^2+1)}^3}dx$

applying reduction formula we get,

$=\frac{x}{4{(x^2+1)}^2}+\frac{3}{4}\int \frac{1}{{(x^2+1)}^2}dx=\frac{3{\tan }^{-1}\left(x\right)}{8}+\frac{3x}{8(x^2+1)}+\frac{x}{4{(x^2+1)}^2}\therefore \int \frac{1}{x^2+1}dx-2\int \frac{1}{{(x^2+1)}^2}dx+\int \frac{1}{{(x^2+1)}^3}dx=\frac{3{\tan }^{-1}\left(x\right)}{8}-\frac{5x}{8(x^2+1)}+\frac{x}{4{(x^2+1)}^2}$

Hence,

$\int \frac{x^{4}dx}{{(x^2+1)}^3}=\frac{3{\tan }^{-1}\left(x\right)}{8}-\frac{5x}{8(x^2+1)}+\frac{x}{4{(x^2+1)}^2}+C=\frac{3{\tan }^{-1}\left(x\right)}{8}+\frac{-5x^3-3x}{8{(x^2+1)}^2}+C.$