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

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

Vedclass · Accepted Answer

$2 \sqrt{e^x-1} - 2 	an^{-1} \sqrt{e^x-1} + c$

Vedclass · Answer

(C) Let $I = \int \sqrt{e^x-1} \, dx$. Substitute $e^x - 1 = t^2$,which implies $e^x = t^2 + 1$. Differentiating both sides,$e^x \, dx = 2t \, dt$. Thus,$dx = \frac{2t}{t^2+1} \, dt$. Substituting these into the integral: $I = \int t \cdot \frac{2t}{t^2+1} \, dt = \int \frac{2t^2}{t^2+1} \, dt$. Rewrite the numerator: $I = \int \frac{2(t^2+1) - 2}{t^2+1} \, dt = \int \left( 2 - \frac{2}{t^2+1} ight) \, dt$. Integrating term by term: $I = 2t - 2 	an^{-1}(t) + c$. Substituting $t = \sqrt{e^x-1}$ back: $I = 2 \sqrt{e^x-1} - 2 	an^{-1} \sqrt{e^x-1} + c$.

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

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