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?
 A: Because the domain of $\ln$. it will not allow negative numbers inside ;) 
A: Consider what happens if $x<1$: The original function makes complete sense, so there ought to be an antiderivative, but your expression doesn't make sense. Calculate it, and $\ln(1-x)$ appears. The two antiderivatives for $x<1$ and $x>1$ are most easily stitched together with an absolute value.
A: A lot of people are telling you it's a domain error, but I will try to give an explanation for a beginner level calculus student looking at this problem. This isn't exactly your problem, but explains why we include the absolute value around $\ln|x|$.
Look at this integral 
$$
\int_{-2}^{-1} \dfrac{1}{x} dx
$$
Now if you graph this function it is very clear that this function is defined on this interval, so you should be able to take the area under the curve right?
Since you know that the antiderivative of $\frac{1}{x}$ is $\ln(x)$, the result should be 
$$
\ln(x)|_{-2}^{-1} = \ln(-1) - \ln(-2)
$$
Wait, that's illegal! $\ln(x)$ is not defined for $x\leq 0$ (well unless we're in complex plane, but thats a whole other can of worms)! So in order to make sure this integral is correctly evaluated for $x<0$, we need to place absolute value signs around the argument
$$
\int_{-2}^{-1} \dfrac{1}{x} dx = \ln|-1| - \ln|-2| = -\ln2
$$
A: Hint: Since we have $$\int\frac{1}{x}dx=ln(|x|)+C$$
