$\mathrm{I}=\frac{1}{2} \int \frac{2 x-2}{x^2-2 x-5} d x+4 \int \frac{d x}{x^2-2 x-5}$
$=\frac{1}{2} \mathrm{I}_1+4 \mathrm{I}_2+\mathrm{C}$ (say)
Put $x^2-2 \mathrm{x}-5=t$, so that $(2 x-2) d x=d t$
$\therefore \quad \mathrm{I}_1=\int \frac{d t}{t}=\log |t|=\log \left|x^2-2 x-5\right|$
$\mathrm{I}_2=\int \frac{d x}{(x-1)^2-\left(\sqrt{6)}^2\right.}=\frac{1}{2 \sqrt{6}} \log \left|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\right|$
$\therefore \mathrm{I}=\frac{1}{2} \log \left|x^2-2 x-5\right|+\frac{2}{\sqrt{6}} \log \left|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\right|+\mathrm{C}$