Question

If $\int log(\sqrt{1-x}+\sqrt{1+x})dx=xf\left(x\right)+Ax+Bsi{n}^{-1}+x+C$, then $f\left(x\right)=log(\sqrt{1-x}+\sqrt{1+x})$ and

• A. $A=-\frac{1}{2},B=\frac{1}{2}$
• B. $A=-\frac{1}{3}$
• C. $B=\frac{2}{3}$
• D. $B=-\frac{1}{2}$

Answer 1

The correct option is A $A=-\frac{1}{2},B=\frac{1}{2}$

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

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

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

$I=xf\left(x\right)-\frac{1}{2}\int \left\{x\left[\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right]\left(\frac{1}{\sqrt{1-x^2}}\right)\right\}dx$

$I=xf\left(x\right)+\frac{1}{2}\int \left(\frac{x}{\sqrt{1-x^2}}\right)\frac{{(\sqrt{1+x}-\sqrt{1-x})}^2}{[(1+x)-(1-x)]}dx$

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

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

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

$I=xf\left(x\right)-\frac{1}{2}\left(x\right)+\frac{1}{2}{\sin }^{-1}x$