int sqrt{e^x-4} , dx equals | vedclass

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(B) Let $I = \int \sqrt{e^x-4} \, dx$. Put $e^x-4 = t^2$. Then $e^x \, dx = 2t \, dt$, which implies $dx = \frac{2t}{e^x} \, dt = \frac{2t}{t^2+4} \,

Vedclass · Accepted Answer

$2 \sqrt{e^x-4}-4 	an ^{-1}\left(\frac{\sqrt{e^x-4}}{2}ight)+C$

Vedclass · Answer

(B) Let $I = \int \sqrt{e^x-4} \, dx$. Put $e^x-4 = t^2$. Then $e^x \, dx = 2t \, dt$, which implies $dx = \frac{2t}{e^x} \, dt = \frac{2t}{t^2+4} \, dt$. Substituting these into the integral: $I = \int t \cdot \frac{2t}{t^2+4} \, dt = 2 \int \frac{t^2}{t^2+4} \, dt$. Now, rewrite the integrand: $I = 2 \int \frac{t^2+4-4}{t^2+4} \, dt = 2 \int \left(1 - \frac{4}{t^2+2^2}ight) \, dt$. Integrating term by term: $I = 2 \left[ t - 4 \cdot \frac{1}{2} 	an^{-1}\left(\frac{t}{2}ight) ight] + C$. $I = 2t - 4 	an^{-1}\left(\frac{t}{2}ight) + C$. Substituting $t = \sqrt{e^x-4}$ back: $I = 2 \sqrt{e^x-4} - 4 	an^{-1}\left(\frac{\sqrt{e^x-4}}{2}ight) + C$.

$\int \sqrt{e^x-4} \, dx$ equals

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