Question

$\int \left(3x-2\right)\sqrt{x^2+x+1}dx=$

Answer 1

$\begin{array}{c}I=\int \left(3x-2\right)\sqrt{x^2+x+1}dx\\ I=\int \left[\frac{3}{2}\left(2x+1\right)-\frac{7}{2}\right]\sqrt{x^2+x+1}dx\\ I=\int \frac{3}{2}\left(2x+1\right)\sqrt{x^2+x+1}dx-\int \frac{7}{2}\sqrt{x^2+x+1}dx\\ Solving,\int \frac{3}{2}\left(2x+1\right)\sqrt{x^2+x+1}dx\\ Let,x^2+x+1=u\\ 2x+1=du\\ \int \frac{3}{2}\left(2x+1\right)\sqrt{x^2+x+1}dx=\frac{3}{2}\int \sqrt{u}du\\ =\frac{3}{2}.\frac{2u^{3/2}}{3}=u^{3/2}={\left(x^2+x+1\right)}^{3/2}\\ Solving,-\int \frac{7}{2}\sqrt{x^2+x+1}dx\\ -\int \frac{7}{2}\sqrt{x^2+x+1}dx=-\int \frac{7}{2}\sqrt{{\left(x^2+\frac{1}{2}\right)}^2+\frac{3}{4}}dx\\ Let,2x+1=u\\ dx=\frac{du}{2}\\ -\int \frac{7}{2}\sqrt{{\left(x^2+\frac{1}{2}\right)}^2+\frac{3}{4}}dx=-\frac{7}{2}\int \sqrt{u^2+1}du\\ =-\frac{7}{2}\left[\frac{3}{2}\ln \sqrt{\frac{u^2}{3}+1}+\frac{u}{\sqrt{3}}\right]+\left[\frac{3\sqrt{u}\sqrt{\frac{u^2}{3}+1}}{2}\right]\\ I={\left(x^2+x+1\right)}^{3/2}-\frac{7}{2}\left[\frac{3}{2}\ln \sqrt{\frac{{\left(2x+1\right)}^2}{3}+1}+\frac{\left(2x+1\right)}{\sqrt{3}}\right]+\left[\frac{3\sqrt{\left(2x+1\right)}\sqrt{\frac{{\left(2x+1\right)}^2}{3}+1}}{2}\right]+C\end{array}$