The value of the integral $\int\limits_{0}^{\log 5}\frac{e^x\sqrt{e^x-1}}{e^x+3}dx$, is 

$4-π$

Putting $e^x-1=t^2$ in the given integral, we have

$I=\int\limits_{0}^{\log 5}\frac{e^x\sqrt{e^x-1}}{e^x+3}dx=2\int\limits_{0}^{2}\frac{t^2}{t^2+4}dt=2\left\{\int\limits_{0}^{2}1\,dt-4\int\limits_{0}^{2}\frac{dt}{t^2+4}\right\}$

$⇒I=2\left[t-2\tan^{-1}(\frac{t}{2})\right]_{0}^{2}=2[2-2×\frac{π}{4}]=4-π$