Answer
$$\int^{\ln\sqrt3}_{0}\frac{e^xdx}{1+e^{2x}}=\frac{\pi}{12}$$
Work Step by Step
$$A=\int^{\ln\sqrt3}_{0}\frac{e^xdx}{1+e^{2x}}$$
We set $u=e^x$, which means $$du=e^xdx$$
For $x=0$, we have $$u=e^0=1$$
For $x=\ln\sqrt3$, we have $$u=e^{\ln\sqrt3}=\sqrt3$$
Therefore, $$A=\int_1^{\sqrt3}\frac{1}{1+u^2}du=\tan^{-1}u\Big]_1^{\sqrt3}$$ $$A=(\tan^{-1}\sqrt3-\tan^{-1}1)$$ $$A=\frac{\pi}{3}-\frac{\pi}{4}=\frac{\pi}{12}$$
