I think I'm getting the hang of evaluating integrals, but I just wanted to check my answer. Is someone able to confirm if this working and result are correct?

$\displaystyle \int^e_1 \frac{ln(x)}{x} dx = \int^e_1 \frac{1}{x}ln(x) dx$

Let

$\displaystyle u = ln(x)$

$\displaystyle du = \frac {1}{x} dx$

Substituting values gives:

$\displaystyle \int^e_1 u du = \left [ \frac{u^2}{2} \right ]^e_1 = \left [ \frac{(ln(x))^2}{2} \right ]^e_1 = \frac {(ln(e))^2}{2} - \frac {(ln(1))^2}{2} = \frac {1}{2} - 0 = \frac{1}{2}$