Question

${\int }_1^4\frac{x^2+x}{\sqrt{2x+1}}dx$

Answer 1

${\int }_1^4\frac{x^2+x}{\sqrt{2x+1}}dx={\int }_1^4\frac{x^2}{\sqrt{2x+1}}+\frac{x}{\sqrt{2x+1}}$

$\frac{1}{2}\left[{\int }_1^4\frac{{\left(\frac{U-1}{2}\right)}^2+\left(\frac{U-1}{2}\right)}{\sqrt{U}}\right]$

$\frac{1}{8}{\int }_1^4\frac{U^2-2U+1+U-1}{\sqrt{U}}$

$\Rightarrow \frac{1}{8}{\int }_1^4{U}^{3/4}-2U^{1/2}+1\frac{1}{8}{\int }_1^4{U}^{3/4}-U^{1/2}du$

$\Rightarrow \frac{1}{8}[\frac{4U^{7/4}}{7}-\frac{2U^{3/2}}{3}]=\frac{1}{8}{\left[\frac{4}{7}\right(2x+1{)}^{7/4}-\frac{2}{3}(2x+1{)}^{3/2}]}_1^4$

$\Rightarrow \frac{1}{8}[\left[\frac{4}{7}\right[9{]}^{7/4}-\frac{2}{3}\left(9{)}^{3/2}\right]-\left[\frac{4}{7}\right(3{)}^{7/4}-\frac{2}{3}\left(3{)}^{3/2}\right]]$

$\Rightarrow 2x+1=U$

$x=\left(\frac{U-1}{2}\right)$

$\frac{2dx}{du}=1$