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?

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.
