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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
up vote 9 down vote accepted

Your solution is an antiderivative of the original function. You can always check whether your solution is correct by taking its derivative. This also implies that the book solution and your solution differ by a constant.

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.

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