# How to integrate function 1/(x^2+a) ?

1. Sep 30, 2011

### Holali

Hi,
I would like to calculate one integral. I just want to get primitive function, not definite integral.

∫ 1/(x^2+a) dx

where a is real number and >0

I only found that

∫ 1/(x^2+1) dx ,
its arctan(x) + C,
but i dont know how it is with different 'a' values.
Thanks for help!

2. Sep 30, 2011

### HallsofIvy

First, this should not have been posted under "differential equations". I will move it to "Calculus and Analysis".

Second, let $y= x/\sqrt{a}$ so that $x= \sqrt{a}y$ and $dx= \sqrt{a}dy$. The integral becomes
$$\int\frac{1}{ay^2+ a}(\sqrt{a}dy)= \frac{\sqrt{a}}{a}\int \frac{1}{y^2+1}dy$$.