How to solve $\int \frac{2x^2-4x+3}{(x-1)^2}\, dx$? My book says the answer to $\int \frac{2x^2-4x+3}{(x-1)^2} \, dx$ is $2x-\frac{1}{x-1}+C$ but symbolab says it is $2x-\frac{2}{x-1}+\frac{4}{x-1}-\frac{3}{x-1}-2+C$. Who is correct and how would I get to the answer? I tried to use u-substitution but that doesn't work because the derivative of $2x^2-4x+3$ is not a multiple of $(x-1)^{-2}$ and the derivative of $x-1$ is not a multiple of $2x^2-4x+3$.

Note that $ \int \frac{2x^2-4x+3}{(x-1)^2}\, dx = \int \frac{2\left(x-1\right)^2+1}{\left(x-1\right)^2}\, dx = \int 2+\frac{1}{\left(x-1\right)^2}\, dx = 2x -  \frac{1}{\left(x-1\right)} + C$ 
