intlimits_0^{ln 5} frac{e^x sqrt{e^x - 1}}{e^ | vedclass

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(A) Let $I = \int \frac{e^x \sqrt{e^x - 1}}{e^x + 3} dx$. Substitute $t = e^x$,then $dt = e^x dx$. The integral becomes $I = \int \frac{\sqrt{t - 1}}{

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$4 - \pi$

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(A) Let $I = \int \frac{e^x \sqrt{e^x - 1}}{e^x + 3} dx$. Substitute $t = e^x$,then $dt = e^x dx$. The integral becomes $I = \int \frac{\sqrt{t - 1}}{t + 3} dt$. Substitute $u = \sqrt{t - 1}$,then $u^2 = t - 1$,so $t = u^2 + 1$ and $2u du = dt$. $I = \int \frac{u \cdot 2u du}{u^2 + 1 + 3} = 2 \int \frac{u^2}{u^2 + 4} du$. $I = 2 \int \left( 1 - \frac{4}{u^2 + 4} ight) du = 2u - 8 \int \frac{1}{u^2 + 2^2} du$. $I = 2u - 8 \cdot \frac{1}{2} 	an^{-1}\left( \frac{u}{2} ight) = 2u - 4 	an^{-1}\left( \frac{u}{2} ight)$. Substituting back $u = \sqrt{e^x - 1}$,we get $2 \sqrt{e^x - 1} - 4 	an^{-1}\left( \frac{\sqrt{e^x - 1}}{2} ight)$. Evaluating from $0$ to $\ln 5$: At $x = \ln 5$,$e^x = 5$,so $u = \sqrt{5 - 1} = 2$. At $x = 0$,$e^x = 1$,so $u = \sqrt{1 - 1} = 0$. Result $= [2(2) - 4 	an^{-1}(1)] - [2(0) - 4 	an^{-1}(0)] = 4 - 4(\frac{\pi}{4}) = 4 - \pi$.

$\int\limits_0^{\ln 5} \frac{e^x \sqrt{e^x - 1}}{e^x + 3} dx = $

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