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?

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 :
$$I=\dfrac{1}{a^2\sqrt{\frac{x}{x-a^2}}}+C$$
