<s>[INST] 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?
 [/INST] 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:
$$=\frac{1}{a^2}sin(u)+c=\frac{1}{a^2}sin(arctan(\frac xa))+c=\frac{x}{a^2 \sqrt{a^2+x^2}}+c$$
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