$\displaystyle \int ln(x^2-4)dx=xln(x^2-4)-\int \frac{x(2x)}{x^2-4}dx$
$\displaystyle =xln(x^2-4)-(xln(x^2-4)+\int ln(x^2-4)dx$
$\displaystyle =\int ln(x^2-4)dx$
Where did i go wrong?
...
You have $\displaystyle I=x\ln{(x^2-4)}-\int \frac{2x^2}{x^2-4}dx$
$\displaystyle \begin{align*} &=x\ln{(x^2-4)}-2\int \frac{x^2-4+4}{x^2-4}dx = x\ln{(x^2-4)}-2{\left[ x+\int \frac{4}{x^2-4}dx\right]} \end{align*}$
$\displaystyle = x\ln{(x^2-4)}-2{\left[ x+\int \frac{1}{x-2}dx-\int\frac{1}{x+2}dx\right]}$
Proceed