Question

How do you test for convergence for `int (lnx)/x^2dx` from 1 to infinity?

Answer 1

The indefinite integral is

#int (lnx)/x^2dx=-int 1/x*lnxdx=-1/x*lnx+int 1/x*1/x*dx= -1/x*lnx-1/x+c#

Hence

`int_1^oo lnx/x^2dx=[-1/x*lnx-1/x]_1^oo=1`

So the intergal converges