Evaluate the following integral:
$\int \frac{1}{(x-1) \sqrt{x^{2}+1}} d x$
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 $\mathrm{t}=\frac{1}{\mathrm{x}-1}$
$-\frac{1}{\sqrt{2}} \log \left(\frac{1}{x-1}+\frac{1}{2}+\sqrt{\left(\frac{1}{x-1}+\frac{1}{2}\right)^{2}+\frac{1}{4}}\right)+c$
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