Question

How do you integrate `1/(x^2+4)`?

Answer 1

`1/2arctan(x/2)+C`

Explanation:

Our goal should be to make this mirror the arctangent integral:

`int1/(u^2+1)du=arctan(u)+C`

To get the `1` in the denominator, start by factoring:

`int1/(x^2+4)dx=int1/(4(x^2/4+1))dx=1/4int1/(x^2/4+1)dx`

Note that we want `u^2=x^2/4`, so we let `u=x/2`, which implies that `du=1/2dx`.

`1/4int1/(x^2/4+1)dx=1/2int(1/2)/((x/2)^2+1)dx=1/2int1/(u^2+1)du`

This is the arctangent integral:

`1/2int1/(u^2+1)du=1/2arctan(u)+C=1/2arctan(x/2)+C`