Question

How do you integrate `int lnx/x^2` by integration by parts method?

Answer 1

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

Explanation:

We seek:

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

We can then apply Integration By Parts:

Let # { (u,=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 \ (lnx)(1/x^2) \ dx = (lnx)(-1/x) - int \ (-1/x)(1/x) \ dx`

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

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

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