Question

Integrate `1/(4+x^2) dx` ?

Answer 1

`int1/(4+x^2)"d"x=1/2arctan(1/2x)+"c"`

Explanation:

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

Let `u=x/2` and `"d"u=1/2"d"x`

Then

`1/4int1/(1+(x/2)^2)"d"x=1/2int1/(1+u^2)"d"u`

This is a well-known integral which evaluates to

`1/2arctanu=1/2arctan(1/2x)+"c"`