Evaluate $ \int ^2 _1 \frac{\ln x}{x^2} \text{dx}$ $\displaystyle \int ^2 _1 \dfrac{\ln x}{x^2} \text{dx}$
My working so far:
$u=\ln x, v'=x^2$
$u'=x^{-1}, v=\dfrac{1}{3}x^3$
Integrating by parts:
$\displaystyle \dfrac{1}{3}x^3\ln x-\int\dfrac{1}{3}x^3\cdot x^{-1}$
$\displaystyle \left[\dfrac{1}{3}x^3\ln x-\int\dfrac{1}{3}x^2\right]^2 _1$
$\left[\dfrac{1}{3}x^3\ln x-\dfrac{1}{9}x^3\right]^2_1$
Is this right so far? The answer is $\frac{1}{2}(1-\ln2)$ but from the integral I got I don't think I can get this answer.

You should have $v'=x^{-2}$, so $v=-x^{-1}$, now go from there.
