Question

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

• A. $2-\pi /2$
• B. $1-\pi /2$
• C. $\pi /2$
• D. $2+\pi /2$

Answer 1

Answer: (A)

$2-\pi /2$

Let $I=\int \frac{\sqrt{x}}{1+x}dx$
Substitute $t=\sqrt{x}\Rightarrow dt=\frac{1}{2\sqrt{x}}ds$
$I=2\int \frac{t^2}{t^2+1}dt=2\int (1-\frac{1}{t^2+1})dt=2\int dt-2\int \frac{1}{t^2+1}dt=2t-2{tan}^{-1}t=2\sqrt{x}-2{tan}^{-1}\sqrt{x}$
Therefore
${\int }_0^{1}Idx={[2\sqrt{x}-2{tan}^{-1}\sqrt{x}]}_0^1=2-\frac{\pi }{2}$