Question:
Evaluate the following integral:
$\int \frac{1}{(x+1) \sqrt{x^{2}+x+1}} d x$
Solution:
assume $x+1=\frac{1}{t}$
$\mathrm{dx}=-\frac{1}{\mathrm{t}^{2}} \mathrm{dt}$
$-\int \frac{d t}{\sqrt{1+t-t^{2}}}$
$-\int \frac{d t}{\sqrt{\frac{5}{4}-\left(t-\frac{1}{2}\right)^{2}}}$
Using identity $\int \frac{\mathrm{dx}}{\sqrt{\mathrm{a}^{2}-\mathrm{x}^{2}}}=\arcsin \left(\frac{\mathrm{x}}{\mathrm{a}}\right)+\mathrm{c}$
$-\arcsin \left(\frac{\left(t-\frac{1}{2}\right)}{\frac{\sqrt{5}}{2}}\right)+c$
Substituting $\mathrm{t}=\frac{1}{\mathrm{x}+1}$
$-\arcsin \left(\frac{\left(\frac{1}{x+1}-\frac{1}{2}\right)}{\frac{\sqrt{5}}{2}}\right)+c$