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Question Number 28885 by abdo imad last updated on 31/Jan/18

$$find{\int }_{-1}^1\frac{dt}{t+\sqrt{1+t^2}}.$$

Commented by abdo imad last updated on 02/Feb/18

$$letputI={\int }_{-1}^1\frac{dt}{t+\sqrt{1+t^2}}$$$$I={\int }_{-1}^1\frac{\sqrt{1+t^2}-t}{1+t^2-t^2}dt={\int }_{-1}^1\sqrt{1+t^{2}dt}-{\int }_{-1}^{1}tdt$$$$=2{\int }_0^1\sqrt{1+t^2}dt-0=2{\int }_0^1\sqrt{1+t^2}dtthech.t=tanxgive$$$${\int }_0^1\sqrt{1+t^2}dt={\int }_0^{\frac{\pi }{4}}cosx(1+{tan}^{2}x)dx={\int }_0^{\frac{\pi }{4}}\frac{dx}{cosx}andthech.$$$$tan\left(\frac{x}{2}\right)=ugive$$$${\int }_0^{\frac{\pi }{4}}\frac{dx}{cosx}={\int }_0^{\sqrt{2}-1}\frac{1}{\frac{1-u^2}{1+u^2}}\frac{2du}{1+u^2}={\int }_0^{\sqrt{2}-1}\frac{2du}{1-u^2}$$$$={\int }_0^{\sqrt{2}-1}(\frac{1}{1+u}+\frac{1}{1-u})du={[ln\mid \frac{1+u}{1-u}\mid ]}_0^{\sqrt{2}-1}$$$$=ln\left(\frac{\sqrt{2}}{2-\sqrt{2}}\right)=ln\left(\frac{\sqrt{2}}{\sqrt{2}(\sqrt{2}-1)}\right)=-ln(\sqrt{2}-1).$$$$I=-2ln(\sqrt{2}-1).$$

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