Đặt: $t=e^x \Rightarrow dt=e^xdx$
Ta có:
$\int\limits_0^1\dfrac{\sqrt{e^x}}{\sqrt{e^x+e^{-x}}}dx$
$=\int\limits_0^1\dfrac{e^x}{\sqrt{e^{2x}+1}}dx$
$=\int\limits_1^e\dfrac{dt}{\sqrt{t^2+1}}$
$=\ln(t+\sqrt{1+t^2})\left|\begin{array}{l}e\\1\end{array}\right.$
$=\ln\dfrac{e+\sqrt{1+e^2}}{1+\sqrt2}$