Question

If $\int \log (\sqrt{1-x}+\sqrt{1+x})dx=xf\left(x\right)+Ax+B{\sin }^{-1}x+c$, then

• A. $f\left(x\right)=\log (\sqrt{1-x}+\sqrt{1+x})$
• B. $A=-\frac{1}{3}$
• C. $B=\frac{2}{3}$
• D. $B=-\frac{1}{2}$

Answer 1

The correct option is A $f\left(x\right)=\log (\sqrt{1-x}+\sqrt{1+x})$

Let, $I=\int \log (\sqrt{1-x}+\sqrt{1+x})dx$

Now using by parts taking $1$ as first function

$I=x\log (\sqrt{1-x}+\sqrt{1+x})-\int \frac{\frac{1}{2\sqrt{1+x}}-\frac{1}{2\sqrt{1-x}}}{\sqrt{1-x}+\sqrt{1+x}}\left(x\right)dx$

Now, $\int \frac{\frac{1}{2\sqrt{1+x}}-\frac{1}{2\sqrt{1-x}}}{\sqrt{1-x}+\sqrt{1+x}}dx=\frac{1}{4}\int \frac{\sqrt{1-x}-\sqrt{1+x}}{\sqrt{1-x}+\sqrt{1+x}}\cdot \frac{x}{\sqrt{1-x^2}}dx$

$=\frac{1}{4}\int \frac{(\sqrt{1-x}-\sqrt{1+x}{)}^2}{(1-x)-(1+x)}\cdot \frac{x}{\sqrt{1-x^2}}dx$, rationalizing first term.

$=\frac{1}{4}\int \frac{\left(1-x\right)+(1+x)-2\sqrt{1-x^2}}{-2x}\cdot \frac{x}{\sqrt{1-x^2}}dx$

$=\frac{1}{4}\int \frac{1-\sqrt{1-x^2}}{\sqrt{1-x^2}}dx=\frac{1}{4}{\sin }^{-1}x-\frac{1}{4}x+c$

Hence, $I=x\log (\sqrt{1-x}+\sqrt{1+x})-\frac{1}{4}{\sin }^{-1}x+\frac{1}{4}x+c$