Question

How do you integrate `int dx/(4x^2+9)^2` using trig substitutions?

Answer 1

`int dx/(4x^2+9)^2=1/108arctan((2x)/3)+x/(72x^2+162)+C`

Explanation:

`int dx/(4x^2+9)^2`

=`1/2int (2dx)/((2x)^2+3^2)^2`

After using `2x=3tanu` and `2dx=3(secu)^2*du` transforms, this integral became

`1/2int (3(secu)^2*du)/(81(secu)^4)`

=`1/54int (cosu)^2*du`

=`1/108int (1+cos2u)*du`

=`1/108u+1/216sin2u+C`

=`1/108u+1/216*(2tanu)/((tanu)^2+1)+C`

After using `2x=3tanu`, `tanu=(2x)/3` and `u=arctan((2x)/3)` inverse transforms, I found

`int dx/(4x^2+9)^2`

=`1/108arctan((2x)/3)+1/216*(2*(2x)/3)/(((2x)/3)^2+1)+C`

=`1/108arctan((2x)/3)+x/(72x^2+162)+C`