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Question Number 31517 by abdo imad last updated on 09/Mar/18

$$find{\int }_{-1}^1\frac{dx}{\sqrt{1+x}+\sqrt{1-x}}.$$

Commented by abdo imad last updated on 12/Mar/18

$$letputI\left(\xi \right)={\int }_{-1+\xi }^{1+\xi }\frac{dx}{\sqrt{1+x}+\sqrt{1-x}}$$$$wehaveI={lim}_{\xi \to 0}I\left(\xi \right)but$$$$I\left(\xi \right)={\int }_{-1+\xi }^{1+\xi }\frac{\sqrt{1+x}-\sqrt{1-x}}{2x}dx$$$$=\frac{1}{2}({\int }_{-1+\xi }^{1+\xi }\frac{\sqrt{1+x}}{x}dx-{\int }_{-1+\xi }^{1+\xi }\frac{\sqrt{1-x}}{x}dx)$$$$ch.\sqrt{1+x}=t\Rightarrow 1+x=t^2\Rightarrow x=t^2-1give$$$${\int }_{-1+\xi }^{1+\xi }\frac{\sqrt{1+x}}{x}dx={\int }_{\sqrt{\xi }}^{\sqrt{2+\xi }}\frac{t}{t^2-1}\left(2t\right)dt$$$$=2{\int }_{\sqrt{\xi }}^{\sqrt{2+\xi }}\frac{t^2-1+1}{t^2-1}dt=2(\sqrt{2+\xi }-\sqrt{\xi })+{\int }_{\sqrt{\xi }}^{\sqrt{2+\xi }}(\frac{1}{t-1}-\frac{1}{t+1})dt$$$$=2(\sqrt{2+\xi }-\sqrt{\xi })+{[ln\mid \frac{t-1}{t+1}\mid ]}_{\sqrt{\xi }}^{\sqrt{2+\xi }}$$$$=2(\sqrt{2+\xi }-\sqrt{\xi })+ln\mid \frac{\sqrt{2+\xi }-1}{\sqrt{2+\xi }+1}\mid -ln\mid \frac{\sqrt{\xi }-1}{\sqrt{\xi }+1}\mid$$$${\to }_{\xi \to 0}2\sqrt{2}+ln\left(\frac{\sqrt{2}-1}{\sqrt{2}+1}\right)andch.\sqrt{1-x}=t\Rightarrow 1-x=t^2\Rightarrow x=1-t^{2}give$$$${\int }_{-1+\xi }^{1+\xi }\frac{\sqrt{1-x}}{x}dx={\int }_{\sqrt{2-\xi }}^{\sqrt{-\xi }}\frac{t}{1-t^2}(-2t)dt$$$$=2{\int }_{\sqrt{2-\xi }}^{\sqrt{-\xi }}\frac{t^2-1+1}{t^2-1}dt=2(\sqrt{-\xi }-\sqrt{2-\xi })+{\int }_{\sqrt{2-\xi }}^{\sqrt{-\xi }}(\frac{1}{t-1}-\frac{1}{t+1})dt$$$$=2(\sqrt{-\xi }-\sqrt{2-\xi })+{[ln\mid \frac{t-1}{t+1}\mid ]}_{\sqrt{2-\xi }}^{\sqrt{-\xi }}$$$$=2(\sqrt{-\xi }-\sqrt{2-\xi })+ln\mid \frac{\sqrt{-\xi }-1}{\sqrt{-\xi }+1}\mid -ln\mid \frac{\sqrt{2-\xi }-1}{\sqrt{2-\xi }+1}\mid$$$$\to -2\sqrt{2}-ln\left(\frac{\sqrt{2}-1}{\sqrt{2}+1}\right)\Rightarrow$$$${lim}_{\xi \to 0}I\left(\xi \right)=\frac{1}{2}(2\sqrt{2}+ln\left(\frac{\sqrt{2}-1}{\sqrt{2}+1}\right)+2\sqrt{2}+ln\left(\frac{\sqrt{2}-1}{\sqrt{2}+1}\right))$$$$=2\sqrt{2}+ln\left(\frac{\sqrt{2}-1}{\sqrt{2}+1}\right).$$

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