Question

What is `int (1+lnx)/x^2dx`?

Answer 1

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

Explanation:

We seek:

`I = int \ (1+lnx)/x^2 \ dx`

In preparation for an application of [integration by Parts]We can then apply Integration By Parts:

Let # { (u,=1+lnx, => (du)/dx,=1/x), ((dv)/dx,=1/x^2, => v,=-1/x ) :}#

Then plugging into the IBP formula:

`int \ (u)((dv)/dx) \ dx = (u)(v) - int \ (v)((du)/dx) \ dx`

We have:

`int \ (1+lnx)(1/x^2) \ dx = (1+lnx)(-1/x) - int \ (-1/x)(1/x) \ dx`

Giving is:

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

`\ \ = -(1+lnx)/x -1/x + C`

`\ \ = -(2+lnx)/x C`