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?
Your partial fraction decomposition is incorrect. It should be $$\dfrac{x+1}{x^2(x-1)} = - \dfrac1{x^2} - \dfrac2x + \dfrac2{x-1}$$ Hence, \begin{align} \int_{-2}^{-1}\dfrac{x+1}{x^2(x-1)} dx & = - \int_{-2}^{-1}\dfrac{dx}{x^2} - \int_{-2}^{-1}\dfrac{2dx}x + \int_{-2}^{-1}\dfrac{2dx}{x-1}\\ & = \left. \left(\dfrac1x - 2 \log \vert x \vert + 2 \log (\vert x-1 \vert) \right) \right \vert_{-2}^{-1}\\ & = \left( -1 - 0 + 2 \log 2\right) - \left( -\dfrac12 - 2\log2 + 2 \log 3\right)\\ & = -\dfrac12 + 4 \log 2 - 2 \log 3\\ & = -\dfrac12 + 2 \left(2 \log 2 - \log 3 \right)\\ & = -\dfrac12 + 2 \log(4/3)\\ \end{align}
Another idea: to have "nicer limits" and avoid absolute values and disgusting stuff like that, do first substitute $\,u=-x\,\,,\,\,du=-dx\,$ , so:
$$\int_{-2}^{-1}\frac{x+1}{x^2(x-1)}dx=\int_2^1\frac{-u+1}{u^2(-u-1)}(-du)=\int_1^2\frac{u-1}{u^2(u+1)}du=$$
$$=-2\int_1^2\frac{du}{u+1}+2\int_1^2\frac{du}{u}-\int_1^2\frac{du}{u^2}=$$
$$\left.\left(-2\log(u+1)+2\log u+\frac{1}{u}\right)\right|_1^2=-2\log\frac{3}{2}+2\log 2-\frac{1}{2}=2\log\frac{4}{3}-\frac{1}{2}$$
