Question

${\int }_0^1\frac{x}{\sqrt{1+x^2}}dx$

Answer 1

${\int }_0^1\frac{xdx}{\sqrt{1+x^2}}$

$=\frac{1}{2}{\int }_0^1\frac{2xdx}{\sqrt{1+x^2}}$

$=\frac{1}{2}{\int }_0^1{(1+x^2)}^{\frac{-1}{2}}$

$=\frac{1}{2}{\left[\frac{{(1+x^2)}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]}_0^1$

$=\frac{1}{2}{\left[\frac{{(1+x^2)}^{\frac{3}{2}}}{\frac{3}{2}}\right]}_0^1$

$=\frac{1}{2}{\left[\frac{2{(1+x^2)}^{\frac{3}{2}}}{3}\right]}_0^1$

$=\frac{1}{2}[\frac{2{(1+1^2)}^{\frac{3}{2}}}{3}-\frac{2{(1+0)}^{\frac{3}{2}}}{3}]$

$=\frac{1}{2}[\frac{2^{\frac{5}{2}}}{3}-\frac{2}{3}]$

$=\frac{1}{6}\left(2(2^{\frac{3}{2}}-1)\right)$

$=\frac{2^{\frac{3}{2}}-1}{3}$

$=\frac{2\sqrt{2}-1}{3}$