Question

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

Answer 1

$$\begin{gathered} {\int }_2^4\frac{x^2+x}{\sqrt{2x+1}}dx \\ ={\left[\left(x^2+x\right)\sqrt{2x+1}\right]}_2^4-{\int }_2^4\left(2x+1\right)\sqrt{2x+1}dx \\ ={\left[\left(x^2+x\right)\sqrt{2x+1}\right]}_2^4-{\int }_2^4{\left(2x+1\right)}^{\frac{3}{2}}dx \\ ={\left[\left(x^2+x\right)\sqrt{2x+1}\right]}_2^4-{\left[\frac{{\left(2x+1\right)}^{{\displaystyle \frac{5}{2}}}}{5}\right]}_2^4 \\ =60-6\sqrt{5}-\left(\frac{243}{5}-\frac{25\sqrt{5}}{5}\right) \\ =\frac{57}{5}-\sqrt{5} \end{gathered}$$