Question

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

Answer 1

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

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

let $x^2+2x-3=t^2$

$\Rightarrow 2tdt=2(x+1)dx\Rightarrow tdt=(x+1)dx$

$\therefore$ $\int \frac{1}{{(x+1)}^4.\sqrt{t^2}}.tdt$

$=\int \frac{tdt}{t{(x^2+1+2x)}^2}$

$=\int \frac{dt}{{(x^2+2x-3+4)}^2}$

$=\int \frac{dt}{{(t^2+4)}^2}$

let $t=2\tan \theta \Rightarrow dt=2{\sec }^2\theta d\theta$

$=\int \frac{2{\sec }^2\theta d\theta }{16{(1+{\tan }^2\theta )}^2}$

$=\frac{1}{8}\int \frac{{\sec }^2\theta d\theta }{{\sec }^2\theta }(\because {\sec }^2\theta -{\tan }^2\theta =1)$

$=\frac{1}{8}\int {\cos }^2\theta d\theta$

$=\frac{1}{8}\int \frac{2{\cos }^2\theta }{2}d\theta =\frac{1}{16}\int (1+\cos 2\theta )d\theta (\because \cos 2\theta =2{\cos }^2\theta -1)$

$=\frac{1}{16}\theta +\frac{\sin 2\theta }{32}$

$=\frac{1}{16}{\tan }^{-1}\left(\frac{\sqrt{x^2+2x-3}}{2}\right)+\frac{4}{32}\left[\frac{\sqrt{(x^2+2x-3)}}{x^2+2x-2}\right]$