Integral of $\frac{x}{(1-x^3)\sqrt{1-x^2}}$ In the pool of difficult (at least to me) integrals I've been trying to solve this one:
$$\int\frac{x}{(1-x^3)\sqrt{1-x^2}}dx$$
Since Wolfram Alpha has been helpful with all the other integrals ( at least I can verify that my solution is correct), I turned to it this time as well. But the result seemed quite ...odd to me. My attemp in solving this was using the substitution $x=\sin{u}$ which leaves me with the following integral:
$$\int\frac{\sin{u}}{1-\sin^3{u}}du$$
But this is as far as I could get . And again, even for this integral, WA returnes something which isn't likely to be a solution to an integral in introductory calculus course.
 A: The quadratic under the radical term $\sqrt{1-x^{2}}=\sqrt{\left(1-x\right)\left(1+x\right)}$ possesses two real roots, which suggests that an Euler substitution of the third kind may be the way to go. Consider the substitution relation,
$$\sqrt{\frac{1+x}{1-x}}=t\implies x=\frac{t^{2}-1}{t^{2}+1};~~~\small{-1<x<1}.$$
Taking differentials of both sides, we obtain
$$\frac{\mathrm{d}x}{\left(1-x\right)^{2}\sqrt{\frac{1+x}{1-x}}}=\mathrm{d}t,$$
or equivalently,
$$\frac{\mathrm{d}x}{\left(1-x\right)\sqrt{1-x^{2}}}=\mathrm{d}t.$$
Now, your problem could be recast as the definite integral,
$$\int_{0}^{z}\frac{x}{\left(1-x^{3}\right)\sqrt{1-x^{2}}}\,\mathrm{d}x;~~~\small{-1<z<1}.$$
Transforming the integral using the substitution rule above yields an integral with rational integrand: for $-1<z<1$,
$$\begin{align}
\int_{0}^{z}\frac{x}{\left(1-x^{3}\right)\sqrt{1-x^{2}}}\,\mathrm{d}x
&=\int_{0}^{z}\frac{x}{\left(1+x+x^{2}\right)\left(1-x\right)\sqrt{1-x^{2}}}\,\mathrm{d}x\\
&=\int_{1}^{\sqrt{\frac{1+z}{1-z}}}\frac{t^{4}-1}{3t^{4}+1}\,\mathrm{d}t.\\
\end{align}$$
From there, you would of course proceed by partial fractions and integrate term by term in the usual manner. Can you take it from here?

A: The substitution $u = \sqrt{\dfrac{1-x}{1+x}}$ gives you 
$$ \int \dfrac{u^4-1}{u^6+3u^2}\; du$$
which you can do with partial fractions: the factorization
$$ u^4 + 3 = \left(u^2 + \sqrt{2} \sqrt[4]{3}\; u + \sqrt{3}\right) \left(u^2  - \sqrt{2} \sqrt[4]{3}\; u + \sqrt{3}\right)$$
may be useful.
