According to the solution manual: $\int \frac{x}{\sqrt{1-x^{4}}}dx = \frac{1}{2}\arcsin x^{2}+C$
My solution doesn't seem to be working. I know another way of solving it (setting $u=x^{2}$) but the fact that this way of solving it doesn't work bothers me.
$$\text{set }u=1-x^{4}\text{ so } dx=\frac{du}{-4x^{3}} $$
$$ \begin{align*} \int \frac{x}{\sqrt{1-x^{4}}}dx &= \int \frac{x}{\sqrt{u}}dx \\ &= \int \frac{xdu}{-4x^{3}\sqrt{u}} \\ &= -\frac{1}{4} \int \frac{du}{x^{2}\sqrt{u}} \\ \end{align*} $$
$$ \text{set } v=\sqrt{u} \text{ so }du=2\sqrt{u}\,dv $$
\begin{align*} -\frac{1}{4} \int \frac{du}{x^{2}\sqrt{u}} &= -\frac{1}{2} \int \frac{dv}{x^{2}} \\ &= -\frac{1}{2} \int \frac{dv}{\sqrt{1-v^{2}}} \\ &= -\frac{1}{2} \arcsin (v) + C \\ &= -\frac{1}{2} \arcsin (\sqrt {1-x^{4}}) + C \\ \end{align*}
I'll be happy to clarify any steps I took. Thanks!