Question

How do you find the integral of `int ln x/(x^3) dx` from 1 to infinity?

Answer 1

The answer is `=1/4`

Explanation:

First , calculate the indefinite integral

`I=int(lnxdx)/(x^3)`

Perform the integration by parts

`intuv'=uv-intu'v`

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

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

Therefore,

`I=-lnx/(2x^2)-int(-(1dx)/(2x^3))`

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

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

Now, calculate the definite intergal

`int_1^t(lnxdx)/(x^3)=[-lnx/(2x^2)-1/(4x^2)]_1^t`

`=(-lnt/(2t^2)-1/(4t^2))-(-ln1/(2*1^2)-1/(4*1^2))`

`=1/4-lnt/(2t^2)-1/(4t^2)`

`lim_(t->oo)-lnt/(2t^2)=0`

`lim_(t->oo)-1/(4t^2)=0`

And finally the integral is

`int_1^ oo(lnxdx)/(x^3)= lim_(t->oo)(1/4-lnt/(2t^2)-1/(4t^2))`

`=1/4`