Question

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

Answer 1

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

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

Let $t=2x^2+1\Rightarrow dt=4xdx$

$=\frac{1}{4}{\int }_1^2\frac{dt}{\sqrt{t}}$

$=\frac{1}{4}{\int }_1^2{t}^{\frac{-1}{2}}dt$

$=\frac{1}{4}{\left[\frac{t^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}\right]}_1^2$

$=\frac{2}{4}{\left[\sqrt{t}\right]}_1^2$

$=\frac{1}{2}(\sqrt{2}-1)$