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Question Number 65015 by aliesam last updated on 24/Jul/19

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

Commented by mathmax by abdo last updated on 24/Jul/19

$$letI=\int \frac{\sqrt{x+1}-\sqrt{x-1}}{\sqrt{x+1}+\sqrt{x-1}}dx\Rightarrow I=\int \frac{{(\sqrt{x+1}-\sqrt{x-1})}^2}{x+1-x+1}dx$$$$=\frac{1}{2}\int (x+1-2\sqrt{x^2-1}+x-1)dx$$$$=\frac{1}{2}\int (2x-2\sqrt{x^2-1})dx=\int xdx-\int \sqrt{x^2-1}dx$$$$=\frac{x^2}{2}-\int \sqrt{x^2-1}dxchang.x=chtgive$$$$\int \sqrt{x^2-1}dx=\int shtshtdt=\int {sh}^{2}tdt=\int \frac{ch\left(2t\right)-1}{2}dt$$$$=\frac{1}{4}sh\left(2t\right)-\frac{t}{2}=\frac{1}{2}shtcht-\frac{t}{2}butt=argch\left(x\right)=ln(x+\sqrt{x^2-1})$$$$\Rightarrow \int \sqrt{x^2-1}dx=\frac{1}{2}x\sqrt{x^2-1}-\frac{1}{2}ln(x+\sqrt{x^2-1})\Rightarrow$$$$I=\frac{x^2}{2}+\frac{1}{2}ln(x+\sqrt{x^2-1})-\frac{1}{2}x\sqrt{x^2-1}+C.$$

Commented by aliesam last updated on 24/Jul/19

$$thankyousir$$

Commented by mathmax by abdo last updated on 24/Jul/19

$$youarewelcome.$$

Answered by Tanmay chaudhury last updated on 24/Jul/19

$$\int \frac{x+1-2\sqrt{x^2-1}+x-1}{x+1-x+1}dx$$$$\int x-\sqrt{x^2-1}dx$$$$\frac{x^2}{2}-(\frac{x\sqrt{x^2-1}}{2}-\frac{1}{2}ln(x+\sqrt{x^2-1})+c$$$$$$

Commented by aliesam last updated on 24/Jul/19

$$thankyousir$$

Commented by Tanmay chaudhury last updated on 24/Jul/19

$$mostwelcomesir$$

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