# Can't Solve an Integral

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!

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Set $x \equiv \sqrt{t}$. – Felix Marin Jan 21 '14 at 7:16

For this specific problem, imagine the right triangle with sides $x^2$ and $\sqrt{1-x^4}$ and hypotenuse $1$. Then $\arcsin\sqrt{1-x^4} = \frac{\pi}{2} - \arcsin x^2$, and it should be easy to see from there how both solutions are related.