I'm having a bit of difficulty with this one. Can anyone help?

Many thanks.

Q.$\displaystyle \int^2_1$$\displaystyle \frac{1}{x^2}$ log_{e}x dx

Attempt:u = log_{e}x => du = $\displaystyle \frac{1}{x}$ dx

dv = $\displaystyle \frac{1}{x^2}$ => dv = x^{-2}=> v = $\displaystyle \int$x^{-2}=> $\displaystyle \frac{-x^-^2}{2}$

$\displaystyle \int$u dv = uv - $\displaystyle \int$v du => $\displaystyle \int$$\displaystyle \frac{1}{x^2}$log_{e}x dx => log_{e}x$\displaystyle (\frac{-x^2}{2})$ + $\displaystyle \int$$\displaystyle \frac{x^-^2}{2}$$\displaystyle (\frac{1}{x})$ => $\displaystyle \frac{-x^-^2log_ex}{2}$ + $\displaystyle \int$$\displaystyle \frac{x^-^3}{2}$ => $\displaystyle \frac{-x^-^2log_ex}{2}$ - $\displaystyle \frac{x^-^2}{4}$

$\displaystyle \int^2_1$$\displaystyle \frac{1}{x^2}$log_{e}x dx = $\displaystyle [\frac{-x^-^2log_ex}{2} - \frac{x^-^2}{4}]^2_1$ => $\displaystyle [\frac{-2^-^2log_e2}{2} - \frac{2^-^2}{4}] - [\frac{-1^-^2log_e1}{2} - \frac{1^-^2}{4}$ => $\displaystyle \frac{-log_e2}{8} - \frac{1}{16} + 0 + \frac{1}{4}$ => $\displaystyle \frac{-log_e2}{8} - \frac{5}{16}$(From text book): $\displaystyle \frac{1}{2}$(1 - ln2)

Ans.: