Question

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

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

Answer 1

Answer: (A), (B)

$f\left(x\right)=\log (\sqrt{1-x}+\sqrt{1+x})$
$L=-\frac{1}{2}$

Let $I=\int \log (\sqrt{1-x}+\sqrt{1+x})dx$
$=x\log (\sqrt{1-x}+\sqrt{1+x})-\int \frac{x}{(\sqrt{1-x}+\sqrt{1+x})}(\frac{-1}{2\sqrt{1-x}}+\frac{1}{2\sqrt{1+x}})dx$
$=x\log (\sqrt{1-x}+\sqrt{1+x})+\frac{1}{2}{I}_1$
Where $I_1=\int \frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\frac{x}{\sqrt{1-x^2}}dx$
$=\int \frac{{(\sqrt{1+x}-\sqrt{1-x})}^2}{1+x-1+x}\frac{x}{\sqrt{1-x^2}}dx$
$=\frac{1}{2}\int \frac{1+x+1-x-2\sqrt{1-x^2}}{\sqrt{1-x^2}}dx=\int (\frac{1}{\sqrt{1-x^2}}-1)dx={\sin }^{-1}x-x$
Therefore $I=x\log (\sqrt{1-x}+\sqrt{1+x})+\frac{1}{2}({\sin }^{-1}x-x)+c$