Question

$\int \frac{x}{\sqrt{x^2+x+1}}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{xdx}{\sqrt{x^2+x+1}} \\ \mathrm{Consider}, \\ x=A\frac{d}{dx}\left(x^2+x+1\right)+B \\ \Rightarrow x=A\left(2x+1\right)+B \\ \Rightarrow x=\left(2A\right)x+A+B \\ \mathrm{Equating}\mathrm{Coefficient}\mathrm{of}\mathrm{like}\mathrm{terms} \\ 2A=1 \\ \Rightarrow A=\frac{1}{2} \\ \mathrm{And} \\ A+B=0 \\ \Rightarrow \frac{1}{2}+B=0 \\ \Rightarrow B=-\frac{1}{2} \\ \therefore I=\int \frac{\left({\displaystyle \frac{1}{2}\left(2x+1\right)-\frac{1}{2}}\right)}{\sqrt{x^2+x+1}}dx \\ =\frac{1}{2}\int \left(\frac{2x+1}{\sqrt{x^2+x+1}}\right)dx-\frac{1}{2}\int \frac{dx}{\sqrt{x^2+x+{\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{4}}+1}} \\ \mathrm{Putting}{x}^2+x+1=t \\ \Rightarrow \left(2x+1\right)dx=dt \\ \mathrm{Then}, \\ I=\frac{1}{2}\int \frac{dt}{\sqrt{t}}-\frac{1}{2}\int \frac{dx}{\sqrt{{\left(x+{\displaystyle \frac{1}{2}}\right)}^2+{\left({\displaystyle \frac{\sqrt{3}}{2}}\right)}^2}} \\ =\frac{1}{2}\int t^{-\frac{1}{2}}dt-\frac{1}{2}\log \left|x+\frac{1}{2}+\sqrt{{\left(x+\frac{1}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}\right|+C \\ =\frac{1}{2}\left|\frac{t^{-{\displaystyle \frac{1}{2}}+1}}{-{\displaystyle \frac{1}{2}}+1}\right|-\frac{1}{2}\log \left|x+\frac{1}{2}+\sqrt{x^2+x+1}\right|+C \\ =\sqrt{t}-\frac{1}{2}\log \left|x+\frac{1}{2}+\sqrt{x^2+x+1}\right|+C \\ =\sqrt{x^2+x+1}-\frac{1}{2}\log \left|x+\frac{1}{2}+\sqrt{x^2+x+1}\right|+C \end{gathered}$$