Integrate: $\int (x^2+a^2)^{-3/2} \cdot dx$ 
Integrate: $\int (x^2+a^2)^{-3/2} \cdot dx$     

My Approach: 
$\int (x^2+a^2)^{-3/2} \cdot dx$
$\int (x^2+a^2)^{-3/2} \cdot d(a^2+x^2)\cdot \frac{dx}{d(x^2+a^2)}$     
But this doesn't give the right answer. I showed this to my friend and he said $d(x^2+a^2)$ is not possible which makes sense since you can't take a small element of the  form $(x^2+a^2)$. How can I then solve this integration without using trigonometry?
 A: Here is the correct solution. Note for anything of the form $({x^2+a^2)^{n/2}}$ for n odd, consider using the following trigonometric substituion.
So, let $x=atan(u) \ (or \ x=asinh(u)$ works aswell$)$ $\Rightarrow dx=(a)sec^2(u)du$
Thus we have:
$$\int \frac{1}{(x^2+a^2)^{3/2}}dx=\int \frac {asec^2u}{(a^2tan^2(u)+a^2)^{3/2}}du=\frac {a}{a^3}\int \frac {sec^2u}{(tan^2(u)+1)^{3/2}}du$$
$$=\frac {1}{a^2}\int \frac {sec^2u}{(tan^2(u)+1)^{3/2}}du=\frac {1}{a^2}\int \frac {sec^2u}{(sec^2(u))^{3/2}}du=\frac {1}{a^2}\int \frac {1}{sec(u)}du=\frac {1}{a^2}\int cos(u)du$$
Finally, we have:
$$=\frac{1}{a^2}sin(u)+c=\frac{1}{a^2}sin(arctan(\frac xa))+c=\frac{x}{a^2 \sqrt{a^2+x^2}}+c$$
A: Following your idea we get $du=2xdx$ so :
$$I=\int (x^2+a^2)^{-3/2}  dx=\int\frac{2xdx}{2x(x^2+a^2)^{3/2}}=\int \frac{du}{2\sqrt{u^4-a^2u^3}}=\frac{1}{2}\int\frac{du}{u^{3/2}\sqrt{u-a^2}} $$
Let $v=\frac{1}{u-a^2}$, we get $dv=-\frac{1}{(u-a^2)^2}$ :
$$I=-\frac{1}{2}{\displaystyle\int}\dfrac{1}{\left(a^2v+1\right)^\frac{3}{2}}\,\mathrm{d}v$$
Let $w=a^2v+1$ then $dw=a^2$ :
$$I=-\frac{1}{2}\class{steps-node}{\cssId{steps-node-2}{\dfrac{1}{a^2}}}{\displaystyle\int}\dfrac{1}{w^\frac{3}{2}}\,\mathrm{d}w=\dfrac{1}{a^2\sqrt{w}}$$
And finally :
$$I=\dfrac{1}{a^2\sqrt{\frac{x}{x-a^2}}}+C$$
A: We use a trick to get a reduction formula:
$$ \frac{1}{(x^2+a^2)^{3/2}} = \frac{1}{a^2} \frac{x^2+a^2-x^2}{(x^2+a^2)^{3/2}} = \frac{1}{a^2} \frac{1}{\sqrt{x^2+a^2}} - \frac{1}{a^2}\frac{x^2}{(x^2+a^2)^{3/2}}. $$
The first term is lower-order, the second can be integrated by parts:
$$ \int \frac{x^2}{(x^2+a^2)^{3/2}} \, dx = -\frac{x}{\sqrt{x^2+a^2}} + \int \frac{dx}{\sqrt{x^2+a^2}}. $$
Oh, but now the new integral on the right here cancels with the integral of the first fraction above! Hence
$$ \int \frac{dx}{(x^2+a^2)^{3/2}} = \frac{1}{a^2} \frac{x}{\sqrt{x^2+a^2}} +C, $$
and it's easy to verify this by differentiating.
