Question

please help
definite integral from 1 to 2 lnx/x^2dx

Answer 1

integration by parts I believe...

Answer 2

\[u=\ln x\]\[dv=x^{-2}\]

Answer 3

yep that will work

du = 1/x
v = -1/x

\[\rightarrow \int\limits_{?}^{?}\frac {\ln x}{x^{2}}dx = -\frac{\ln x}{x} +\int\limits_{?}^{?}\frac{1}{x^{2}} dx\]