Question

How do you integrate `int tan^-1x` by integration by parts method?

Answer 1

The integral `=xarctanx-ln(1+x^2)/2+C`

Explanation:

The integration by parts is `intu'v=uv-intuv'`
Here we have `u'=1` `=>` `u=x`
and `v=arctanx` `=>` `v'=1/(1+x^2)`
If `v=arctanx` `=>` `x=tanv`
So by differentiating, `1=(1/cos^2v)((dv)/dx)`
`v'=(dv)/dx=cos^2v=1/(1+x^2)`

The integration
`intarctanxdx=xarctanx-int(xdx)/(1+x^2)`
`int(xdx)/(1+x^2)=(1/2)int(2xdx)/(1+x^2)`
Let `u=1+x^2` `=>` `du=2xdx`
`int(2xdx)/(1+x^2)=int(du)/u=lnu`
`int(xdx)/(1+x^2)=1/2ln(1+x^2)`

And finally `intarctanxdx=xarctanx-ln(1+x^2)/2+C`