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Question Number 31516 by abdo imad last updated on 09/Mar/18

$$find\int \frac{dx}{x+\sqrt{1+x^2}}.$$

Commented by abdo imad last updated on 16/Mar/18

$$I=\int (\sqrt{1+x^2}-x)dx=\int \sqrt{1+x^2}dx-\frac{x^2}{2}ch.x=sht$$$$give\int \sqrt{1+x^2}dx=\int cht.chtdt=\int {ch}^{2}tdt$$$$=\int \frac{1+ch\left(2t\right)}{2}dt=\frac{t}{2}+\frac{1}{2}\int ch\left(2t\right)dt$$$$=\frac{t}{2}+\frac{1}{4}sh\left(2t\right)=\frac{t}{2}+\frac{1}{2}sh\left(t\right)ch\left(t\right)$$$$=\frac{1}{2}argshx+\frac{1}{2}argsh\left(x\right)\sqrt{1+x^2}$$$$=\frac{1}{2}ln(x+\sqrt{1+x^2})+\frac{\sqrt{1+x^2}}{2}ln(x+\sqrt{1+x^2})\Rightarrow$$$$I=\frac{1}{2}(1+\sqrt{1+x^2})ln(x+\sqrt{1+x^2})-\frac{x^2}{2}+\lambda$$

Answered by math1967 last updated on 18/Mar/18

$$\int \frac{x-\sqrt{1+x^2}}{x^2-1-x^2}dx$$$$\int \sqrt{1+x^2}dx-\int xdx$$$$\frac{x\sqrt{1+x^2}}{2}+\frac{1}{2}log[x+\sqrt{1+x^2}]-\frac{x^2}{2}+C$$

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