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Question Number 125897 by john_santu last updated on 15/Dec/20

$$\int \frac{dx}{2+x+\sqrt{1-x^2}}?$$

Commented by MJS_new last updated on 15/Dec/20

$$\mathrm{the}\mathrm{path}\mathrm{is}$$$$t=\frac{1+\sqrt{1-x^2}}{x}\iff x=\frac{2t}{t^2+1}\to dx=-\frac{x^2\sqrt{1-x^2}}{1+\sqrt{1-x^2}}$$$$\mathrm{then}\mathrm{decompose}\mathrm{and}\mathrm{solve}$$

Commented by liberty last updated on 15/Dec/20

$$I=\frac{1}{2}\arcsin x+\frac{1}{2}\ln \mid 2+x+\sqrt{1-x^2}\mid -\sqrt{2}\arctan \left(\frac{\frac{1-\sqrt{1-x^2}}{x}+1}{\sqrt{2}}\right)+c$$$$I=\frac{1}{2}(\arcsin x+\ln \mid 2+x+\sqrt{1-x^2}\mid )-\sqrt{2}\arctan \left(\frac{1+x-\sqrt{1-x^2}}{x\sqrt{2}}\right)+c$$

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