Question

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

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \left(x+1\right)\sqrt{x^2+x+1}dx \\ \mathrm{Also},x+1=\lambda \frac{d}{dx}\left(x^2+x+1\right)+\mu \\ \Rightarrow x+1=\lambda \left(2x+1\right)+\mu \\ \Rightarrow x+1=\left(2\lambda \right)x+\lambda +\mu \\ \mathrm{Equating}\mathrm{coefficient}\mathrm{of}\mathrm{like}\mathrm{terms} \\ 2\lambda =1 \\ \Rightarrow \lambda =\frac{1}{2} \\ \mathrm{And} \\ \lambda +\mu =1 \\ \Rightarrow \frac{1}{2}+\mu =1 \\ \therefore \mu =\frac{1}{2} \\ \therefore I=\frac{1}{2}\int \left(2x+1\right)\sqrt{x^2+x+1}dx+\frac{1}{2}\int \sqrt{x^2+x+1}dx \\ =\frac{1}{2}\int \left(2x+1\right)\sqrt{x^2+x+1}dx+\frac{1}{2}\int \sqrt{x^2+x+{\left(\frac{1}{2}\right)}^2-{\left(\frac{1}{2}\right)}^2+1}dx \\ =\frac{1}{2}\int \left(2x+1\right)\sqrt{x^2+x+1}dx+\frac{1}{2}\int \sqrt{{\left(x+\frac{1}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}dx \\ \mathrm{Let}{x}^2+x+1=t \\ \Rightarrow \left(2x+1\right)dx=dt \\ \mathrm{Then}, \end{gathered}$$
$$\begin{gathered} I=\frac{1}{2}\int \sqrt{t}dt+\frac{1}{2}\left[\frac{x+{\displaystyle \frac{1}{2}}}{2}\sqrt{{\left(x+\frac{1}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}+\frac{3}{8}\log \left|\left(x+\frac{1}{2}\right)+\sqrt{{\left(x+\frac{1}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}\right|\right]+C \\ =\frac{1}{2}\times \frac{2}{3}{t}^{\frac{3}{2}}+\frac{1}{2}\left[\left(\frac{2x+1}{4}\right)\sqrt{x^{\mathit{2}}+x+1}+\frac{3}{8}\log \left|\left(x+\frac{1}{2}\right)+\sqrt{x^2+x+1}\right|\right]+C \\ =\frac{1}{3}{\left(x^2+x+1\right)}^{\frac{3}{2}}+\frac{1}{2}\left[\left(\frac{2x+1}{4}\right)\sqrt{x^2+x+1}+\frac{3}{8}\log \left|\left(x+\frac{1}{2}\right)+\sqrt{x^2+x+1}\right|\right]+C \end{gathered}$$