Evaluate the following integrals:

Let $x^{2}=t$

$2 x d x=d t$ or $x d x=d t / 2$

Hence, $\int \frac{x}{\sqrt{4-x^{4}}}=\int \frac{d t}{2\left(\sqrt{2^{2}-t^{2}}\right)}$

Since we have, $\int \frac{1}{\sqrt{a^{2}-x^{2}}} d x=\sin ^{-1}\left(\frac{x}{a}\right)+c$

So, $\int \frac{\mathrm{dt}}{2\left(\sqrt{2^{2}-\mathrm{t}^{2}}\right)}=\frac{1}{2} \sin ^{-1}\left(\frac{\mathrm{t}}{2}\right)+\mathrm{c}$

Put $t=x^{2}$

$=\frac{1}{2} \sin ^{-1}\left(\frac{t}{2}\right)+c=\frac{1}{2} \sin ^{-1}\left(\frac{x^{2}}{2}\right)+c$