Question

Compute the integral ${\int }_1^0\frac{\sqrt{x}dx}{1+x}$

• A. $-2-\frac{\pi }{2}$
• B. $2+\frac{\pi }{2}$
• C. $\frac{\pi }{2}$
• D. $2$

Answer 1

The correct option is A $-2-\frac{\pi }{2}$

Let $I={\int }_1^0\frac{\sqrt{x}dx}{1+x}$
Put $t=\sqrt{x}\Rightarrow 2tdt=dx$
$I={\int }_1^0\frac{2t^2}{t^2+1}dt=2{\int }_1^0(1-\frac{1}{t^2+1})dt$
$={[-2{\tan }^{-1}t]}_1^0+{\left[2t\right]}_1^0$
$=-2-\frac{\pi }{2}$