How to integrate $\int\frac{dx}{(A^2+x^2)^{3/2}}$ I have the following to integrate:

$$\int\frac{dx}{(A^2+x^2)^{3/2}}$$

Where, $A$ is a constant.

Not really sure what technique to use. Any hint would be helpful.
let $x=A\tan\theta\implies dx=A\sec^2\theta\ d\theta$
$$\int \frac{1}{(A^2+x^2)^{3/2}}dx$$
$$=\int \frac{1}{(A^2+A^2\tan^2\theta)^{3/2}}(A\sec^2\theta\ d\theta)$$
$$=\int \frac{A\sec^2\theta\ d\theta}{A^{3}\sec^3\theta}$$
$$=\frac{1}{A^2}\int \cos \theta\ d\theta$$
$$=\frac{1}{A^2}\sin\theta+C$$
$$=\frac{1}{A^2}\sin\left(\tan^{-1}\frac{x}{A}\right)+C$$
$$=\frac{1}{A^2}\sin\left(\sin^{-1}\frac{x/A}{\sqrt{1+(x/A)^2}}\right)+C$$
$$=\color{red}{\frac{x}{A^2\sqrt{A^2+x^2}}+C}$$
This is actually a simple integral. Use the substitution

$$x = A\tan (y) ~~~~~~~ \text{d}x = A\sec^2(y)\ \text{d}y$$

In this way

$$(A^2 + x^2)^{3/2} = (A^2\tan^2(y) + A^2)^{3/2} = A^3\sec^3(y)$$

and $y = \arctan(x/A)$

So your integration becomes

$$\frac{1}{A^2}\int\cos(y)\ \text{d}y$$

which is trivial.

Final answer:

$$\frac{\sin(\arctan(x/A))}{A^2}$$