<s>[INST] Evaluate $ \int \frac{x^4}{(2-x^2)^{3/2}}dx$ 
Evaluate the following integral:
  $ \int \frac{x^4}{(2-x^2)^{3/2}}dx$

I've tried to apply Chebyshev theorem on the integration of binomial differentials.
We have $ m=4,a=2,b=-1,n=2,p=-3/2$. 
$\frac{m+1}{n}+p$ is integer then we do the substitution
$t^2=2x^{-2}-1$, $x^2=\frac{2}{t^2+1}$
It go me there: $\int\frac{2}{t^2+1}^2(2-\frac{2}{t^2+1})^{-3/2}-\sqrt{2}t(\frac{1}{t^2+1})^{3/2}dt$ which is just more complicated expression. Where I went wrong?
 [/INST] Another way:
Integrate by parts
$$\int\dfrac{x^4}{(2-x^2)^{3/2}}dx=\int x^3\cdot\dfrac x{(2-x^2)^{3/2}}dx$$
$$=x^3\int\dfrac x{(2-x^2)^{3/2}}dx-\int\left[\dfrac{d(x^3)}{dx}\int\dfrac x{(2-x^2)^{3/2}}dx\right]dx$$
Finally, $$\dfrac{x^2}{\sqrt{2-x^2}}=\dfrac{x^2-2+2}{\sqrt{2-x^2}}=\dfrac2{\sqrt{2-x^2}}-\sqrt{2-x^2}$$
Use $\#1,\#8$ of INTEGRALS CONTAINING THE SQUARE ROOT OF $a^2-x^2$
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