Question

The value of the integral ${\int }_0^1(\log (\sqrt{1-x}+\sqrt{1+x})+\left[x\right])dx$ is

• A. $\frac{1}{2}(\log 2-\frac{1}{2}+\frac{\pi }{4})$
• B. $\frac{1}{2}(\log 2-1+\frac{\pi }{2})$
• C. $\frac{1}{3}(\log 4-1+\frac{\pi }{8})$
• D. $\frac{1}{4}(\log 3-1+\frac{\pi }{2})$

Answer 1

The correct option is B $\frac{1}{2}(\log 2-1+\frac{\pi }{2})$

Let $I={\int }_0^1(\log (\sqrt{1-x}+\sqrt{1+x})+\left[x\right])dx$
As $xϵ(0,1)\left[x\right]=0$
$I={\int }_0^1\left(\log (\sqrt{1-x}+\sqrt{1+x})\right)dx$
Integrating by parts by taking 1 as second parts
$I={\left[x\log (\sqrt{1-x}+\sqrt{1+x})\right]}_0^1-\int \frac{\frac{-1}{\sqrt{1-x}}+\frac{1}{\sqrt{1+x}}}{\sqrt{1-x}+\sqrt{1+x}}xdx$
$=\frac{1}{2}\log 2-\frac{1}{2}+\frac{\pi }{4}$