Evaluate the following integrals:

let $I=\int \frac{3 x^{5}}{1+x^{12}} d x$

$=\int \frac{3 x^{5}}{1+\left(x^{6}\right)^{2}} d x$

Let $\mathrm{x}^{6}=\mathrm{t} \ldots . .$ (i)

$\Rightarrow 6 x^{5} d x=d t$

$I=\frac{3}{6} \int \frac{1}{(t)^{2}+1} d t$

$I=\frac{1}{2} \tan ^{-1} t+c$

[since, $\left.\int \frac{1}{1+(\mathrm{x})^{2}} \mathrm{dx}=\tan ^{-1} \mathrm{x}+\mathrm{c}\right]$

$I=\frac{1}{2} \tan ^{-1}\left(x^{6}\right)+c$ [using (i)]