Evaluate the following integral:

assume $x-1=\frac{1}{t}$

$\mathrm{dx}=-\frac{1}{\mathrm{t}^{2}} \mathrm{dt}$

$-\int \frac{d t}{\sqrt{2 t^{2}+2 t+1}}$

$-\frac{1}{\sqrt{2}} \int \frac{d t}{\sqrt{\left(t+\frac{1}{2}\right)^{2}+\frac{1}{4}}}$

Using identity $\int \frac{d x}{\sqrt{x^{2}+a^{2}}}=\log \left(x+\sqrt{x^{2}+a^{2}}\right)+c$

$-\frac{1}{\sqrt{2}} \log \left(\mathrm{t}+\frac{1}{2}+\sqrt{\left.\left(\mathrm{t}+\frac{1}{2}\right)^{2}+\frac{1}{4}\right)}+\mathrm{c}\right.$

Substituting $\mathrm{t}=\frac{1}{\mathrm{x}-1}$

$-\frac{1}{\sqrt{2}} \log \left(\frac{1}{x-1}+\frac{1}{2}+\sqrt{\left(\frac{1}{x-1}+\frac{1}{2}\right)^{2}+\frac{1}{4}}\right)+c$