Finding convergence (or divergence) of $\int_0^{1/e} \frac{dx}{x \ln^2 x}$.

Question

Compute or prove the divergence of $$\int_0^{1/e} \frac{dx}{x \ln^2 x}$$

I have tried to find the usual derivative taking integration from α..(0<α<1/e)to 1/e..it turned out be log0..but the given solution is 1.

Answer 1

HINT

Substitute $u = \ln x$ with $du = dx/x$ to get $\int u^{-2}du$...

Answer 2

$$\int_0^{\frac1e}\dfrac{1}{x\ln^2x}\ dx$$

Use u-substitution: $u=\ln x\implies du=\dfrac{dx}{x}$

first compute without boundaries $$=\int\dfrac{1}{u^2}\ du=\dfrac{u^{-2+1}}{-2+1}=-\left[\dfrac{1}{\ln x}\right]_0^{\frac1e}=1$$