Question

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

• A. $2[\frac{5}{6}+log2]$
• B. $2[\frac{5}{6}-log2]$
• C. $[\frac{5}{6}-log2]$
• D. $[\frac{5}{6}+log2]$

Answer 1

The correct option is B $2[\frac{5}{6}-log2]$

$\sqrt{x}=t\Rightarrow \frac{1}{2\sqrt{x}}dx=dt\Rightarrow dx=2tdt$
$x=0\Rightarrow t=0,x=1\Rightarrow t=1$
${\int }_0^1\frac{x}{1+\sqrt{x}}dx={\int }_0^1\frac{t^2}{1+t}.2tdx=2{\int }_0^1\frac{t^3}{1+t}dt=2{\int }_0^1(1-t+t^2-\frac{1}{1+t})dt$
$=2{[t-\frac{t^2}{2}+\frac{t^3}{2}-log(1+t\left)\right]}_0^1=2[1-\frac{1}{2}+\frac{1}{3}-log2]$
$=2[\frac{5}{6}-log2]$