Evaluate the following integrals:

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 $t=\tan ^{-1} x^{2}$

$\Rightarrow \frac{\left(\tan ^{-1} x^{2}\right)^{2}}{4}+C$

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