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 

$$2x  \color{red}{-\frac{2}{x-1}+\frac{4}{x-1}-\frac{3}{x-1}} \color{green}{-2+K} = 2x \color{red}{-\frac{1}{x-1}} \color{green}{+C}.$$
Well both the answers are correct. The thing is what is asked in the question. The answers are exactly same.