antiderivatives of $\frac1{x-1}$ I was told to find the anti-derivative of this problem: $$\int\frac{2x^2-13x+18}{x-1} dx$$
I solved the problem - first dividing and then finding the anti-derivative: $$x^2-11x+7(\ln(x-1))+c$$
The given answer was the same as mine, but they put the $x-1$ in $\ln(x-1)$ in absolute value - $\ln|x-1|$. Why?

Hint: Since we have $$\int\frac{1}{x}dx=ln(|x|)+C$$
