Chapter 6.2, Problem 108E

To determine

The reason behind the below calculation of the integral being wrong and what will be the correct solution,

$\begin{array}{c}{\displaystyle \int \frac{x}{x^2-1}}dx={\displaystyle \int \frac{x}{u}du}\\ =x{\displaystyle \int \frac{1}{u}du}\\ =x\ln \left|u\right|+C\\ =x\ln \left|x^2-1\right|+C\end{array}$

Where, the substitution for the integral is $u=x^2-1,\frac{du}{dx}=2x\text{ and }dx=\frac{1}{2x}du$.