Question

How do you evaluate the indefinite integral of `dx/(81+x^2)^2`?

Answer 1

Hi, here the polynome is irreducible you can't do partial fraction so we will use substitution

Let's `x=9tan(u)`
`dx = 9sec^2(u) du`
`u=arctan(1/9x)`

So we have :

`int1/(81+x^2)^2dx = int1/(81+81tan^2(u))^2*sec^2(u)du`

`=1/7291int1/(1+tan^2(u))^2*sec^2(u)du`

Don't forget `1+tan^2(u) = sec^2(u)`

`1/729intcos^4(u)*sec^2(u)du = 1/729intcos^2(u)du`

Don't forget `cos^2(u) = 1/2(1+cos(2u))`

`1/1458int1+cos(2u)du = 1/2916int2+2cos(2u)du`

`1/2916[2u+sin(2u)]+C`

`1/2916[2arctan(1/9x)+sin(2arctan(1/9x))]`

`sin(arctan(x)) = x/sqrt(x^2+1)`

so we have

`1/1458[arctan(1/9x)+(9x)/sqrt(x^2+81)]+C`