Question

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

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \left(x+1\right)\sqrt{x^2-x+1}dx \\ \mathrm{Also},x+1=\mathrm{\lambda }\frac{d}{dx}\left(x^2-x+1\right)+\mathrm{\mu } \\ \Rightarrow x+1=\mathrm{\lambda }\left(2x-1\right)+\mathrm{\mu } \\ \Rightarrow x+1=\mathrm{\lambda }\left(2x-1\right)+\mathrm{\mu } \\ \Rightarrow x+1=\left(2\mathrm{\lambda }\right)x+\mathrm{\mu }-\mathrm{\lambda } \\ \mathrm{Equating}\mathrm{the}\mathrm{coefficient}\mathrm{of}\mathrm{like}\mathrm{terms} \\ 2\mathrm{\lambda }=1 \\ \Rightarrow \mathrm{\lambda }=\frac{1}{2} \\ \mathrm{And} \\ \mathrm{\mu }-\mathrm{\lambda }=1 \\ \Rightarrow \mathrm{\mu }-\frac{1}{2}=1 \\ \Rightarrow \mathrm{\mu }=\frac{3}{2} \\ \therefore I=\int \left[\frac{1}{2}\left(2x-1\right)+\frac{3}{2}\right]\sqrt{x^2-x+1}dx \\ =\frac{1}{2}\int \left(2x-1\right)\sqrt{x^2-x+1}dx+\frac{3}{2}\int \sqrt{x^2-x+1}dx \\ =\frac{1}{2}\int \left(2x-1\right)\sqrt{x^2-x+1}dx+\frac{3}{2}\int \sqrt{x^2-x+\frac{1}{4}-\frac{1}{4}+1}dx \\ =\frac{1}{2}\int \left(2x-1\right)\sqrt{x^2-x+1}dx+\frac{3}{2}\int \sqrt{{\left(x-\frac{1}{2}\right)}^2+\frac{3}{4}}dx \\ =\frac{1}{2}\int \left(2x-1\right)\sqrt{x^2-x+1}dx+\frac{3}{2}\int \sqrt{{\left(x-\frac{1}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}dx \\ \mathrm{Let}{x}^2-x+1=t \\ \Rightarrow \left(2x-1\right)dx=dt \\ \therefore I=\frac{1}{2}\int \sqrt{t}dt+\frac{3}{2}\int \sqrt{{\left(x-\frac{1}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}dx \\ =\frac{1}{2}\times \left(\frac{t^{{\displaystyle \frac{3}{2}}}}{{\displaystyle \frac{3}{2}}}\right)+\frac{3}{2}\left[\frac{x-{\displaystyle \frac{1}{2}}}{2}\sqrt{{\left(x-\frac{1}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}+\frac{3}{8}\log \left|x-\frac{1}{2}+\sqrt{{\left(x-\frac{1}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}\right|\right]+\mathrm{C} \\ =\frac{1}{3}{\left(x^2-x+1\right)}^{\frac{3}{2}}+\frac{3}{8}\left(2x-1\right)\sqrt{x^2-x+1}+\frac{9}{16}\log \left|x-\frac{1}{2}+\sqrt{x^2-x+1}\right|+\mathrm{C} \end{gathered}$$