Question

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

Answer 1

`x/(2+2x^2)+arctanx/2+C`

Explanation:

`I=intdx/(1+x^2)^2`

We will use the substitution `x=tantheta`, implying that `dx=sec^2thetad theta`:

`I=int(sec^2thetad theta)/(1+tan^2theta)^2`

Note that `1+tan^2theta=sec^2theta`:

`I=int(sec^2thetad theta)/sec^4theta=int(d theta)/sec^2theta=intcos^2thetad theta`

Recall that `cos2theta=2cos^2theta-1`, so `cos^2theta=1/2cos2theta+1/2`.

`I=1/2intcos2thetad theta+int1/2d theta`

The first integral can be found with substitution (try `u=2theta`).

`I=1/4sin2theta+1/2theta+C`

From `x=tantheta` we see that `theta=arctanx`.

Furthermore, we see that `1/4sin2theta=1/4(2sinthetacostheta)=1/2sinthetacostheta`.

Also, since `tantheta=x`, we can draw a right triangle with the side opposite `theta` being `x`, the adjacent side being `1`, and the hypotenuse being `sqrt(1+x^2)`. Thus, `sintheta=x/sqrt(1+x^2)` and `costheta=1/sqrt(1+x^2)`:

`I=1/2sinthetacostheta+1/2arctanx+C`

`I=1/2(x/sqrt(1+x^2))(1/sqrt(1+x^2))+arctanx/2+C`

`I=x/(2(1+x^2))+arctanx/2+C`