Evaluate $ \int \frac{x^4}{(2-x^2)^{3/2}}dx$

Question

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?

Answer 1

Hint:

Another way:

Use Trigonometric substitution

As $2-x^2\ge0,$ WLOG $x=\sqrt2\sin t$

$\sqrt{2-x^2}=\sqrt2\cos t, dx=?$

Answer 2

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 $a^2-x^2$

Answer 3

$\int \frac{x^4}{(2-x^2)^{3/2}}dx$

Comparing with $\int x^m (a+bx^n)^p dx$

$m=4,a=2,b=-1,n=2,p=-\frac{3}{2}$

Here $\frac{m+1}{n}+p=\frac{5}{2}-\frac{3}{2}=1$, integer.

So we first transform the integral $ x^m (a+bx^n)^p $ by factoring $x^n$ i.e., by $x^2$ in this way

$\int \frac{x^4}{(2-x^2)^{\frac{3}{2}}}dx=\int {x^4}{(2-x^2)^{-\frac{3}{2}}}dx=\int {x^4}{(x^2)^{-\frac{3}{2}}}{(2x^{-2}-1)^{-\frac{3}{2}}}dx=\int x{(2x^{-2}-1)^{-\frac{3}{2}}}dx$

Putting $2x^{-2}-1=z$ give $-4x dx =dz\implies x dx=-\frac{1}{4}dz$

Now $\int \frac{x^4}{(2-x^2)^{\frac{3}{2}}}dx=\int x{(2x^{-2}-1)^{-\frac{3}{2}}}dx=-\frac{1}{4}\int {z^{-\frac{3}{2}}}dz=-\frac{1}{4}\frac{z^{-\frac{1}{2}}}{-\frac{1}{2}}+c=\frac{1}{2} {(2x^{-2}-1)}^{-\frac{1}{2}}+c$