Question

What is the integral of `int(x tan^-1 x)/(1+x^2)^(3/2)` ?

`int(x tan^-1 x)/(1+x^2)^(3/2)`

Answer 1

`(1-tan^-1(x))/sqrt(x^2+1)+C`

Explanation:

`I=int(xtan^-1(x))/(1+x^2)^(3/2)color(red)(dx)`

Let `x=tantheta`, which implies that `dx=sec^2thetad theta`.

Then:

`I=int(tanthetatan^-1(tantheta))/(1+tan^2theta)^(3/2)sec^2thetad theta`

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

`I=int(thetatantheta)/sec^3thetasec^2thetad theta`

`I=intthetatantheta 1/secthetad theta`

`I=intthetasintheta/costhetacosthetad theta`

`I=intthetasinthetad theta`

This looks primed for integration by parts. Let:

`{(u=theta,du=d theta),(dv=sinthetad theta,v=-costheta):}`

Then:

`I=uv-intvdu`

`I=-thetacostheta+intcosthetad theta`

`I=-thetacostheta+sintheta+C`

Remember that `x=tantheta`. Imagine the right triangle associated with this: we must have the leg opposite `theta` being `x` and the adjacent leg being `1`. Then the hypotenuse is `sqrt(x^2+1)`.

Thus `costheta=1/sqrt(x^2+1)` and `sintheta=x/sqrt(x^2+1)`. Moreover, note that `theta=tan^-1(x)`.

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

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