The value of the integral int_0^{log 5} {frac | vedclass

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(B) Let $I = \int_0^{\log 5} {\frac{{{e^x}\sqrt {{e^x} - 1} }}{{{e^x} + 3}}} \,dx$. Substitute ${e^x} - 1 = {t^2}$,so ${e^x} = {t^2} + 1$. Differentia

Vedclass · Accepted Answer

$4 - \pi $

Vedclass · Answer

(B) Let $I = \int_0^{\log 5} {\frac{{{e^x}\sqrt {{e^x} - 1} }}{{{e^x} + 3}}} \,dx$. Substitute ${e^x} - 1 = {t^2}$,so ${e^x} = {t^2} + 1$. Differentiating both sides,${e^x}dx = 2t\,dt$. When $x = 0$,${t^2} = {e^0} - 1 = 0$,so $t = 0$. When $x = \log 5$,${t^2} = {e^{\log 5}} - 1 = 5 - 1 = 4$,so $t = 2$. Substituting these into the integral: $I = \int_0^2 {\frac{{2t \cdot t}}{{{t^2} + 1 + 3}}} \,dt = \int_0^2 {\frac{{2{t^2}}}{{{t^2} + 4}}} \,dt$. $I = 2 \int_0^2 {\frac{{{t^2} + 4 - 4}}{{{t^2} + 4}}} \,dt = 2 \left[ {\int_0^2 {1\,dt - 4\int_0^2 {\frac{{dt}}{{{t^2} + 2^2}}} } } ight]$. $I = 2 \left[ {t|_0^2 - 4 \cdot \frac{1}{2} 	an^{-1}(\frac{t}{2})|_0^2} ight]$. $I = 2 \left[ {2 - 2 	an^{-1}(1)} ight] = 2 \left[ {2 - 2(\frac{\pi}{4})} ight] = 4 - \pi$.

The value of the integral $\int_0^{\log 5} {\frac{{{e^x}\sqrt {{e^x} - 1} }}{{{e^x} + 3}}} \,dx = $

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