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Question Number 96705 by bemath last updated on 04/Jun/20

$$\int \ln (\sqrt{1-\mathrm{x}}+\sqrt{1+\mathrm{x}})\mathrm{dx}=?$$

Answered by john santu last updated on 04/Jun/20

$$\mathrm{D}.\mathrm{I}\mathrm{method}$$$$\mathrm{I}=x\ln (\sqrt{1-x}+\sqrt{1+x})-\int \frac{x(\frac{-1}{2\sqrt{1-x}}+\frac{1}{2\sqrt{1+x}})}{\sqrt{1-x}+\sqrt{1+x}}dx$$$$\mathrm{I}=x\ln (\sqrt{1-x}+\sqrt{1+x})-\frac{1}{2}\int x\left(\frac{\sqrt{1-x^2}-1}{x\sqrt{1-x^2}}\right)dx$$$$\mathrm{I}=x\ln (\sqrt{1-x}+\sqrt{1+x})-\frac{x}{2}+\frac{1}{2}\int \frac{dx}{\sqrt{1-x^2}}$$$$\mathrm{I}=x\ln (\sqrt{1-x}+\sqrt{1+x})-\frac{x}{2}+\frac{1}{2}{\sin }^{-1}\left(x\right)+c$$

Commented by peter frank last updated on 04/Jun/20

$$\mathrm{thank}\mathrm{you}$$

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