Question

$\int \frac{x^2+x+1}{x^2-x}dx$

Answer 1

$$\begin{gathered} \int \left(\frac{x^2+x+1}{x^2-x}\right)dx \\ \frac{x^2+x+1}{x^2-x}=1+\frac{2x+1}{x^2-x} \\ \therefore \int \left(\frac{x^2+x+1}{x^2-x}\right)dx \\ =\int \left(1+\frac{2x+1}{x^2-x}\right)dx \\ =\int 1+\left(\frac{2x-1+2}{x^2-x}\right)dx \\ =\int dx+\int \frac{\left(2x-1\right)dx}{x^2-x}+\int \frac{2dx}{x^2-x+{\left({\displaystyle \frac{1}{2}}\right)}^2-{\left({\displaystyle \frac{1}{2}}\right)}^2} \\ =\int dx+\int \frac{\left(2x-1\right)dx}{x^2-x}+2\int \frac{dx}{{\left(x-{\displaystyle \frac{1}{2}}\right)}^2-{\left({\displaystyle \frac{1}{2}}\right)}^2} \\ =x+\log \left|x^2-x\right|+2\times \frac{1}{2\times {\displaystyle \frac{1}{2}}}\log \left|\frac{x-{\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{2}}}{x-{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{2}}}\right| \\ =x+\log \left|x^2-x\right|+2\log \left|\frac{x-1}{x}\right|+C \end{gathered}$$