How can I evaluate the following integral? We have to calculate the following integral quite often with Gauss surfaces in my Physics course. The teacher never exlained how we can find it and always gives us the result . How can I evaluate it? 
$$\int \frac{dx}{(a^2+x^2)^{3/2}}$$
a is a constant
P.S. In case it's been uploaded I'll delete the question. I couldn't find it after giving a look though.
 A: If you make the substitution $x=a\tan\theta$, then $a^2+x^2=a^2\sec^2\theta$ and $\frac{dx}{d\theta}=a\sec^2\theta$, so the result is
$$ \int\frac{a\sec^2\theta}{a^3\sec^3\theta}\;d\theta=a^{-2}\int \cos\theta\;d\theta=a^{-2}\sin\theta+C=\frac{x}{a^2\sqrt{a^2+x^2}}+C$$
Note that $\sin\theta$ can be determined using $\tan\theta=\frac{x}{a}$ and the Pythagorean theorem.
A: In the first place, "solve" is the wrong word. You are evaluating an integral. One evaluates expressions; one solves equations; one solves problems.
If you see $(a^2+x^2)^{3/2}$ and don't think of the trigonometric substitution $x = a \tan\theta$ then you need to review trigonometric substitutions. Then you have $a^2+x^2 = a^2 (1 + \tan^2\theta) = a^2 \sec^2\theta,$ so $(a^2+x^2)^{3/2} = a^3 \sec^3\theta,$ and $dx = a\sec\theta\tan\theta\,d\theta.$ So you have
$$
\int \frac{dx}{(a^2+x^2)^{3/2}} = \int \frac{a\sec^2\theta\,d\theta}{a^3\sec^3\theta} = \frac 1 {a^2} \int \cos\theta\,d\theta = \cdots\cdots.
$$
Once you get a function of $\theta$ you need to convert it back into a function of $x$, and that will also require remembering some trigonometry.
