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Question Number 200960 by sonukgindia last updated on 27/Nov/23

Answered by Frix last updated on 27/Nov/23

$$t=1+\sqrt{x^2+1}$$$$\Rightarrow$$$$\int ...=\int (1-\frac{1}{t})dt=t-\ln t=$$$$=1+\sqrt{x^2+1}-\ln (1+\sqrt{x^2+1})+C$$$${[1+\sqrt{x^2+1}-\ln (1+\sqrt{x^2+1})]}_{-a}^a=0$$

Answered by Frix last updated on 27/Nov/23

$$=\underset{-1}{\stackrel{1}{\int }}\frac{\sqrt{x^2+1}-1}{x}dx$$$$f\left(x\right)=\frac{\sqrt{x^2+1}-1}{x}$$$$f(-x)=-f\left(x\right)$$

Answered by Sutrisno last updated on 29/Nov/23

$${\int }_{-1}^1\frac{\sqrt{x^2+1}+(x-1)}{\sqrt{x^2+1}+x+1}.\frac{\sqrt{x^2+1}-(x+1)}{\sqrt{x^2+1}-(x+1)}dx$$$${\int }_{-1}^1\frac{\sqrt{x^2+1}-1)}{x}dx$$$$let:\sqrt{x^2+1}=t\to x=\sqrt{t^2-1}\to dx=\frac{t}{x}dt$$$$\int \frac{t-1}{\sqrt{t^2-1}}.\frac{t}{x}dt$$$$\int \frac{t-1}{\sqrt{t^2-1}}.\frac{t}{\sqrt{t^2-1}}dt$$$$\int \frac{t(t-1)}{(t+1)(t-1)}dt$$$$\int \frac{t}{(t+1)}dt$$$$=t-ln(t+1)$$$$=\sqrt{x^2+1}-ln(\sqrt{x^2+1}+1){\mid }_{-1}^1$$$$=\sqrt{2}-ln(\sqrt{2}+1)-(\sqrt{2}-ln(\sqrt{2}+1))$$$$=0$$$$$$

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