Question

What is the antiderivative of `lnx/x^2`?

Answer 1

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

Explanation:

`u=ln x, dv=1/x^2`
`du=1/x, v=-1/x`
`intudv=uv-intvdu`
`=-lnx/x-int-1/x^2 dx`
`=-lnx/x+int1/x^2 dx`
`=-lnx/x-1/x`
`=-(lnx+1)/x`