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

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(C) ધારો કે $I = \int \frac{dx}{(x+1) \sqrt{x^2+1}}$.$x+1 = \frac{1}{t}$ આદેશ લેતા,$dx = -\frac{1}{t^2} dt$ મળે.તેથી $x = \frac{1}{t} - 1 = \frac{1-t}

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

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(C) ધારો કે $I = \int \frac{dx}{(x+1) \sqrt{x^2+1}}$.$x+1 = \frac{1}{t}$ આદેશ લેતા,$dx = -\frac{1}{t^2} dt$ મળે.તેથી $x = \frac{1}{t} - 1 = \frac{1-t}{t}$.હવે $x^2+1 = \left(\frac{1-t}{t}\right)^2 + 1 = \frac{2t^2-2t+1}{t^2}$.આ કિંમતો સંકલનમાં મૂકતા:$I = \int \frac{-dt/t^2}{(1/t) \sqrt{(2t^2-2t+1)/t^2}} = -\int \frac{dt}{\sqrt{2t^2-2t+1}} = -\frac{1}{\sqrt{2}} \int \frac{dt}{\sqrt{(t-1/2)^2 + 1/4}}$.સૂત્ર $\int \frac{du}{\sqrt{u^2+a^2}} = \operatorname{Sinh}^{-1}(\frac{u}{a})$ નો ઉપયોગ કરતા:$I = -\frac{1}{\sqrt{2}} \operatorname{Sinh}^{-1}\left(\frac{t-1/2}{1/2}\right) + c = -\frac{1}{\sqrt{2}} \operatorname{Sinh}^{-1}(2t-1) + c$.$t = \frac{1}{x+1}$ હોવાથી,$2t-1 = \frac{1-x}{x+1}$.તેથી,$I = -\frac{1}{\sqrt{2}} \operatorname{Sinh}^{-1}\left(\frac{1-x}{1+x}\right) + c$.

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

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