I would like to evaluate the following integral $\int_{-2}^{-1} \frac{x+1}{x^2(x-1)} dx$
I tried to solve it by partial fractions as $\int_{-2}^{-1} \left(\frac2x + \frac{-1}{x^2} + \frac{-2}{x-1} \right)dx $ and I got $2\left.\ln{\frac1{x-1}}\right|_{-2}^{-1} $ it to be $2\ln({3\over 2})$. But the right solution is $2\ln \left({4\over 3} \right)-{1\over 2}$. Where did I go wrong?