int frac{dx}{(x+1) sqrt{x^2+1}} = | vedclass

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(C) Let $I = \int \frac{dx}{(x+1) \sqrt{x^2+1}}$.Substitute $x+1 = \frac{1}{t}$,then $dx = -\frac{1}{t^2} dt$.Also,$x = \frac{1}{t} - 1 = \frac{1-t}{t

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$-\frac{1}{\sqrt{2}} \operatorname{Sinh}^{-1}\left(\frac{1+x}{\sqrt{2}(1-x)}\right) + c$

Vedclass · Answer

(C) Let $I = \int \frac{dx}{(x+1) \sqrt{x^2+1}}$.Substitute $x+1 = \frac{1}{t}$,then $dx = -\frac{1}{t^2} dt$.Also,$x = \frac{1}{t} - 1 = \frac{1-t}{t}$.Then $x^2+1 = \left(\frac{1-t}{t}\right)^2 + 1 = \frac{1-2t+t^2+t^2}{t^2} = \frac{2t^2-2t+1}{t^2}$.Substituting these into the integral:$I = \int \frac{-dt/t^2}{(1/t) \sqrt{(2t^2-2t+1)/t^2}} = -\int \frac{dt}{\sqrt{2t^2-2t+1}} = -\int \frac{dt}{\sqrt{2(t^2-t+1/2)}} = -\frac{1}{\sqrt{2}} \int \frac{dt}{\sqrt{(t-1/2)^2 + 1/4}}$.Using the formula $\int \frac{du}{\sqrt{u^2+a^2}} = \operatorname{Sinh}^{-1}(\frac{u}{a})$:$I = -\frac{1}{\sqrt{2}} \operatorname{Sinh}^{-1}\left(\frac{t-1/2}{1/2}\right) + c = -\frac{1}{\sqrt{2}} \operatorname{Sinh}^{-1}(2t-1) + c$.Since $t = \frac{1}{x+1}$,$2t-1 = \frac{2}{x+1} - 1 = \frac{2-x-1}{x+1} = \frac{1-x}{x+1}$.Thus,$I = -\frac{1}{\sqrt{2}} \operatorname{Sinh}^{-1}\left(\frac{1-x}{1+x}\right) + c$.

$\int \frac{dx}{(x+1) \sqrt{x^2+1}} = $

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