1. ## Integral 16

$\int \frac{x^2dx}{\sqrt{(4-x^2)^5}}$

2. I learnt this from Krizalid aka the Inte-Killer. Try the substitution u = 1/x.

I wouldn't be surprised if Kriz has another clever substitution.

3. Trigonometric substitution, $x=2 \sin t$ is a way.

Another way is below.

I assume that $x>0$,

$\frac{x^2}{(4-x^2)^{\frac{5}{2}}} = \frac{x^2}{x^5\left ( \frac{4}{x^2}-1 \right )^{\frac{5}{2}}} = \frac{1}{x^3 \left ( \frac{4}{x^2}-1 \right )^{\frac{5}{2}}}$

$\int \frac{dx}{x^3 \left ( \frac{4}{x^2}-1 \right )^{\frac{5}{2}}}$

Let $u = \sqrt{\frac{4}{x^2}-1}$, then $u^2 = \frac{4}{x^2}-1$ and $2 u~du = -\frac{8}{x^3}~dx$

Applying the substitution gives us $-\frac{1}{4}\int u^{-4}~du= \frac{1}{12u^3}+C = \frac{1}{12 \left ( \frac{4}{x^2}-1 \right )^{\frac{3}{2}}}+C$

For $x < 0$, the integral is $-\frac{1}{12 \left ( \frac{4}{x^2}-1 \right )^{\frac{3}{2}}}+C$.

So we can say that, for any x, the integral is $\frac{\sqrt{x^2}}{12x \left ( \frac{4}{x^2}-1 \right )^{\frac{3}{2}}}+C$

4. I thought I'd elaborate more on my post.

$\int \frac{x^2}{\sqrt{(4-x^2)^5}}dx$

Let $u = \frac{1}{x}$

$du = -\frac{1}{x^2} dx$

$- x^2 du = dx$

Replace:
$\int \frac{1}{u^4\sqrt{(4-\frac{1}{u^2})^5}}du$

$\int \frac{1}{u^4\sqrt{(\frac{4u^2-1}{u^2})^5}}du$

$\int \frac{u^5}{u^4\sqrt{(4u^2-1)^5}}du$

$\int \frac{u}{\sqrt{(4u^2-1)^5}}du$

Now it became a matter of simple substition ( $z = 4u^2 - 1$).

5. Hello, Apprentice123!

Another approach . . .

$\int \frac{x^2\,dx}{\sqrt{(4-x^2)^5}}\;=\;\int\frac{x^2\,dx}{(4-x)^{\frac{5}{2}}}$

Let: $x = 2\sin\theta \quad\Rightarrow\quad dx = 2\cos\theta\,d\theta$

Substitute: . $\int\frac{(2\sin\theta)^2(2\cos\theta\,d\theta)}{( 4\cos^2\!\theta)^{\frac{5}{2}}} \;=\;\int\frac{4\sin^2\!\theta\cdot2\cos\theta\,d\ theta}{32\cos^5\!\theta} \;= \;\frac{1}{4}\int\frac{\sin^2\!\theta}{\cos^4\!\th eta}\,d\theta$

. . $= \;\frac{1}{4}\int \frac{\sin^2\!\theta}{\cos^2\!\theta}\cdot\frac{1} {\cos^2\!\theta}\,d\theta \;=\;\frac{1}{4}\int\tan^2\!\theta(\sec^2\!\theta\ ,d\theta)$

Let $u = \tan\theta \quad\Rightarrow\quad du = \sec^2\!\theta\,d\theta$

Substitute: . $\frac{1}{4}\int u^2\,du \;=\;\frac{1}{12}u^3 + C$

Back-substitute: . $\frac{1}{12}\tan^3\!\theta + C$

We had: . $\sin\theta = \frac{x}{2}\quad\Rightarrow\quad \tan\theta = \frac{x}{\sqrt{4-x^2}}$

Answer: . $\frac{1}{12}\left(\frac{x}{\sqrt{4-x^2}}\right)^3 + C \;=\;\frac{x^3}{12(4-x^2)^{\frac{3}{2}}} + C$