Answer
$$\displaystyle\int_{-1}^{1}\frac{e^{\arctan{y}}}{1+y^{2}}\thinspace dy=e^{\frac{\pi}{4}}-e^{\frac{-\pi}{4}}$$
Work Step by Step
$$\displaystyle\int_{-1}^{1}\frac{e^{\arctan{y}}}{1+y^{2}}\thinspace dy$$
$u= \arctan{y}\quad du=\frac{dy}{1+y^{2}}$
Limits of integration will change from $\displaystyle\int_{-1}^{1}$ to $\displaystyle\int_{\arctan{(-1)}}^{\arctan{(1)}} = \displaystyle\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}$ $$ \displaystyle\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}e^{u}\thinspace du$$ $$=\bigg[e^{u}\bigg]^{\frac{\pi}{4}}_{\frac{-\pi}{4}}$$ $$=e^{\frac{\pi}{4}}-e^{\frac{-\pi}{4}}$$
