Question

$\underset{1}{\overset{4}{\int }}\frac{x^2+x}{\sqrt{2x+1}}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_1^4\frac{x^2+x}{\sqrt{2x+1}}\mathit{}\mathit{d}x\mathit{.}\mathit{} \\ \mathrm{Let}2x+1=u \\ \Rightarrow x=\frac{u-1}{2} \\ \Rightarrow dx=\frac{du}{2} \\ \therefore I=\int \frac{{\left({\displaystyle \frac{u-1}{2}}\right)}^2+{\displaystyle \frac{u-1}{2}}}{\sqrt{u}}\frac{du}{2} \\ \Rightarrow I=\frac{1}{8}\int \frac{u^2+1-2u+2u-2}{\sqrt{u}}du \\ =\frac{1}{8}\int \frac{\left(u^2-1\right)}{\sqrt{u}}du \\ =\frac{1}{8}\int \left(u^{\frac{3}{2}}-u^{-\frac{1}{2}}\right)du \\ =\frac{1}{8}\left[\frac{2u^{{\displaystyle \frac{5}{2}}}}{5}-\frac{2u^{{\displaystyle \frac{1}{2}}}}{1}\right] \\ =\frac{1}{8}{\left[\frac{2{\left(2x+1\right)}^{\frac{5}{2}}}{5}-\frac{2{\left(2x+1\right)}^{{\displaystyle \frac{1}{2}}}}{1}\right]}_1^4 \\ =\frac{1}{8}\left[\frac{2}{5}\times 243-6-\frac{2}{5}\times 9\sqrt{3}+2\sqrt{3}\right] \\ \Rightarrow I=\frac{1}{8}\left[\frac{456}{5}-\frac{8\sqrt{3}}{5}\right] \\ \Rightarrow I=\frac{57-\sqrt{3}}{5} \end{gathered}$$