int {sqrt {{e^x} - 1} } dx = | vedclass

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Question

(A) Let $I = \int {\sqrt {{e^x} - 1} } dx$. Substitute ${e^x} - 1 = {t^2}$,which implies ${e^x} = {t^2} + 1$. Differentiating both sides,we get ${e^x}

Vedclass · Accepted Answer

$2\left[ {\sqrt {{e^x} - 1} - {{	an }^{ - 1}}\sqrt {{e^x} - 1} } ight] + c$

Vedclass · Answer

(A) Let $I = \int {\sqrt {{e^x} - 1} } dx$. Substitute ${e^x} - 1 = {t^2}$,which implies ${e^x} = {t^2} + 1$. Differentiating both sides,we get ${e^x} dx = 2t dt$,so $dx = \frac{2t}{t^2 + 1} dt$. Substituting these into the integral: $I = \int t \cdot \frac{2t}{t^2 + 1} dt = \int \frac{2t^2}{t^2 + 1} dt$. Now,rewrite the integrand: $I = \int \frac{2(t^2 + 1) - 2}{t^2 + 1} dt = \int \left( 2 - \frac{2}{t^2 + 1} ight) dt$. Integrating term by term: $I = 2t - 2 	an^{-1}(t) + c$. Substituting back $t = \sqrt{e^x - 1}$: $I = 2\sqrt{e^x - 1} - 2 	an^{-1}(\sqrt{e^x - 1}) + c = 2\left[ \sqrt{e^x - 1} - 	an^{-1}(\sqrt{e^x - 1}) ight] + c$.

$\int {\sqrt {{e^x} - 1} } dx = $

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