Question

Find `I = int \ lnx/x^2 \ dx`?

Answer 1

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

Explanation:

We seek:

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

We can apply integration by 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`

gives us

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

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

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

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