Answer
$$\int^{\pi/4}_{0}(1+e^{\tan\theta})\sec^2\theta d\theta=e$$
Work Step by Step
$$A=\int^{\pi/4}_{0}(1+e^{\tan\theta})\sec^2\theta d\theta$$
We set $u=\tan\theta$, which means $$du=\sec^2\theta d\theta$$
For $\theta=\pi/4$, we have $$u=\tan(\pi/4)=1$$
For $\theta=0$, we have $$u=\tan0=0$$
Therefore, $$A=\int^1_0(1+e^u)du=\int^1_01du+\int^1_0e^udu$$ $$A=u\Big]^1_0+e^u\Big]^1_0$$ $$A=1+(e-e^0)$$ $$A=1+e-1=e$$
