Question:
Evaluate the following integrals:
$\int \frac{x \tan ^{-1} x^{2}}{1+x^{4}} d x$
Solution:
Assume $\tan ^{-1} x^{2}=t$'
$d\left(\tan ^{-1} x^{2}\right)=d t$
$\Rightarrow \frac{2 x}{x^{4}+1}=d t$
$\Rightarrow \frac{x}{x^{4}+1}=\frac{d t}{2}$
Substituting $\mathrm{t}$ and $\mathrm{dt}$
$\Rightarrow \frac{1}{2} \int \mathrm{tdt}$
$\Rightarrow \frac{t^{2}}{4}+c$
But $t=\tan ^{-1} x^{2}$
$\Rightarrow \frac{\left(\tan ^{-1} x^{2}\right)^{2}}{4}+C$