Question

How do you find the integral of `1/(x^2+4)`?

Answer 1

`1/2tan^(-1)(x/2)+C`

Explanation:

Recall the rule : `int1/(x^2+a^2)dx=1/atan^(-1)(x/a)+C`

We may use this rule to integrate the given function as follows :

`int1/(x^2+4)dx=int1/(x^2+2^2)dx`

`=1/2tan^(-1)(x/2)+C`