Question

Evaluate: $\int \frac{x^2}{1+x^4}dx$

Answer 1

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

Let, $x^2=\tan \theta$.......(1)

$\therefore 2xdx={\sec }^2\theta$

Using these in the above integration we have

$\frac{1}{2}$ $\int \frac{\sqrt{\tan \theta }{\sec }^2\theta }{{\sec }^2\theta }$ $d\theta$

$=\frac{1}{2}\int$ $\sqrt{\tan \theta }d\theta$......(2)

Now,

$\int$ $\sqrt{\tan \theta }d\theta$

$=\frac{1}{2}$ $\int$ $(\{\sqrt{\tan \theta }-\sqrt{\cot \theta }\}+\{\sqrt{\tan \theta }+\sqrt{\cot \theta }\})d\theta$

$=\frac{1}{2}$ $\int$ $(\left\{\frac{\sin \theta -\cos \theta }{\sqrt{\sin \theta \cos \theta }}\right\}+\left\{\frac{\sin \theta +\cos \theta }{\sqrt{\sin \theta \cos \theta }}\right\})d\theta$

$=\frac{1}{2}$ $\int$ $(\left\{\frac{\sin \theta -\cos \theta }{\frac{1}{\sqrt{2}}\sqrt{2\sin \theta \cos \theta }}\right\}+\left\{\frac{\sin \theta +\cos \theta }{\frac{1}{\sqrt{2}}\sqrt{2\sin \theta \cos \theta }}\right\})d\theta$

$=\frac{1}{\sqrt{2}}$ $\int$ $(\left\{\frac{\sin \theta -\cos \theta }{\sqrt{1+2\sin \theta \cos \theta -1}}\right\}+\left\{\frac{\sin \theta +\cos \theta }{\sqrt{1-(1-2\sin \theta \cos \theta )}}\right\})d\theta$

$=\frac{1}{\sqrt{2}}$ $\int$ $(\left\{\frac{-d(\sin \theta +\cos \theta )}{\sqrt{(\sin \theta +\cos \theta {)}^2-1}}\right\}+\left\{\frac{d(\sin \theta -\cos \theta )}{\sqrt{1-(\sin \theta -\cos \theta )}}\right\})$

$=\frac{1}{\sqrt{2}}$ $[-\left\{log|\right(\sin \theta +\cos \theta )+\sqrt{(\sin \theta +\cos \theta {)}^2-1}|\}+{\sin }^{-1}(\sin \theta -\cos \theta \left)\right]$ $+c$

$=\frac{1}{\sqrt{2}}$ $[-\left\{log|\right(\sin \theta +\cos \theta )+\sqrt{2\sin \theta \cos \theta }|\}+{\sin }^{-1}(\sin \theta -\cos \theta \left)\right]$ $+c$

$=\frac{1}{\sqrt{2}}$ $[-\left\{log|\frac{x^2+1}{\sqrt{1+x^4}}+\sqrt{\frac{2x^2}{1+x^4}}|\right\}+{\sin }^{-1}(\frac{x^2-1}{\sqrt{1+x^4}})]$ $+c$ [Since $\sin \theta =\frac{x^2}{\sqrt{1+x^4}}$ and $\cos \theta =\frac{1}{\sqrt{1+x^4}}$ (from 1)]

From (2) the required integral

$=\frac{1}{2\sqrt{2}}$ $[-\left\{log|\frac{x^2+1}{\sqrt{1+x^4}}+\sqrt{\frac{2x^2}{1+x^4}}|\right\}+{\sin }^{-1}(\frac{x^2-1}{\sqrt{1+x^4}})]$ $+c$