Solving $\int \frac{dx}{(x^2 + y^2 + z^2)^{\frac{3}{2}}}$ This was in an old exam in a physics for mathematicians class. I haven't had to deal with these kind of integrals for a while and can't think of a decent substitution. I asked my teacher about it and he mumbled a bit and told me to "google it". 
I've tried a few obvious ones but none of them seemed to work.
Any help would be greatly appreciated.
 A: Integration by parts easily gives:
$$\int\frac{dx}{(x^2+A)^{3/2}} = C+\frac{x}{A\sqrt{x^2+A}}.$$
A: Classic Substitution Method
Let $x=\sqrt{y^2+z^2}\tan \theta$ and thus $dx=\sqrt{y^2+z^2}\sec^2 \theta \,d\theta$.  Then,
$$\begin{align}
\int\frac{dx}{(x^2+y^2+z^2)^{3/2}}&=\frac{1}{y^2+z^2}\int\cos \theta \,d\theta\\\\
&=\frac{1}{y^2+z^2}\sin \theta +C\\\\
&=\frac{1}{y^2+z^2}\frac{x}{\sqrt{x^2+y^2+z^2}}+C
\end{align}$$
A: Since $y,z$ are constant let $\alpha^{2}=y^{2}+z^{2}$. Then let $x=\alpha \sinh t$. $dx = \alpha \cosh t$, so $$\int \frac{1}{(x^{2}+y^{2}+z^{2})^{3/2}}dx=\int\frac{\alpha \cosh t}{(\alpha^{2}\sinh^{2} t + \alpha^{2})^{3/2}}dt\\=\int\frac{\cosh t}{\alpha^{2}\cosh t}dt=\frac{t}{\alpha^{2}}=\frac{1}{\alpha^{2}}\sinh^{-1} \frac{x}{\alpha}$$
A: Following daniel's answer..
$\int \frac{1}{(x^{2}+y^{2}+z^{2})^{3/2}}dx$
With $a^2 = y^2 + z^2$ 
$ \therefore \int \frac{1}{(x^{2}+a^{2})^{3/2}}dx$
Hyperbolic substitutions $x=a\dot{}sinht$ and $dx=a\dot{}cosht$ 
$\int\frac{{a\dot{}cosh(t)}}{(a^2\dot{}sinh^{2}(t)+a^2)^{3/2}}dt$
$cosh^{2}(t)-sinh^{2}(t)=1$
$a^{2}(cosh^2(t)) = a^2(1+sinh^{2}(t)) =  a^2cosh^2(t)=a^2sinh^2(t)+a^2$
Which becomes $\int\frac{{a\dot{}cosh(t)}}{(a^2\dot{}cosh^{2}(t))^{3/2}}dt=\int\frac{{(a\dot{}cosh(t))}}{(a^3\dot{}cosh^{3}(t))}dt=\int\frac{{(a\dot{}cosh(t))}}{(a\dot{}cosh(t))^{3}}dt=\int\frac{{dt}}{(a\dot{}cosh(t))^2}$
Of course I might be wrong but, I think you did something wrong with the exponential fraction part. And as it stands, This is probably even harder than the original question.
I would have loved to comment instead. Apparently I'm not allowed. The answer by Jack is correct.
