Question

How do you integrate `int (lnx)^2/x^3` using integration by parts?

Answer 1

The answer is `=-(lnx)^2/(2x^2)-lnx/(2x^2)-1/(4x^2)+C`

Explanation:

The integration by oarts is

`intuv'dx=uv-intu'vdx`

Here,

`u=(lnx)^2`, `=>`, `u'=2lnx*1/x`

`v'=1/x^3`, `=>`, `v=-1/(2x^2)`

Therefore,

`int((lnx)^2dx)/x^3=-(lnx)^2/(2x^2)-int(-lnxdx)/x^3=-(lnx)^2/(2x^2)+int(lnx)/x^3`

We do once more the integration by parts

`u=lnx`,`=>`, `u'=1/x`

`v'=1/x^3`, `=>`, `v=-1/(2x^2)`

So,

`int(lnx)/x^3=-lnx/(2x^2)+intdx/(2x^3)=-lnx/(2x^2)-1/(4x^2)`

Therefore,

`int((lnx)^2dx)/x^3=-(lnx)^2/(2x^2)-lnx/(2x^2)-1/(4x^2)+C`