Question

$\int \frac{1}{\left(x+1\right)\sqrt{x^2+x+1}}dx$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \frac{dx}{\left(x+1\right)\sqrt{x^2+x+1}} \\ \mathrm{Putting}x+1=\frac{1}{t} \\ \Rightarrow dx=-\frac{1}{t^2}dt \\ \therefore I=\int \frac{-{\displaystyle \frac{1}{t^2}}dt}{{\displaystyle \frac{1}{t}\sqrt{{\left(\frac{1}{t},-,1\right)}^2+-1+1\frac{1}{t}}}} \\ =\int \frac{-{\displaystyle \frac{1}{t^2}}dt}{{\displaystyle \frac{1}{t}\sqrt{\frac{1}{t^2}-+1+\frac{2}{t}\frac{1}{t}}}} \\ =\int \frac{-{\displaystyle \frac{1}{t}}dt}{{\displaystyle \frac{\sqrt{t^2+t-2t+1}}{t}}} \\ =-\int \frac{dt}{\sqrt{t^2-t+1}} \\ =-\int \frac{dt}{\sqrt{t^2-t+{\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{4}}+1}} \\ =-\int \frac{dt}{\sqrt{{\left(t-{\displaystyle \frac{1}{2}}\right)}^2+{\left({\displaystyle \frac{\sqrt{3}}{2}}\right)}^2}} \\ =-\log \left|t-\frac{1}{2}+\sqrt{{\left(t-\frac{1}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}\right|+C \\ =-\log \left|t-\frac{1}{2}+\sqrt{t^2-t+1}\right|+C \\ =-\log \left|\frac{1}{x+1}-\frac{1}{2}+\sqrt{\frac{1}{{\left(x+1\right)}^2}-\frac{1}{x+1}+1}\right|+C \\ =-\log \left|\frac{1}{x+1}-\frac{1}{2}+\frac{\sqrt{{\left(x+1\right)}^2-\left(x+1\right)+1}}{x+1}\right|+C \\ =-\log \left|\frac{1}{x+1}-\frac{1}{2}+\frac{\sqrt{x^2+x+1}}{x+1}\right|+C \end{gathered}$$