Answer
$$e^{\pi/4 }-e^{-\pi/4 }$$
Work Step by Step
Given
$$\int_{-1}^{1} \frac{e^{\tan^{-1}y}}{1+y^2}dy$$
Let $u=\tan^{-1}y\ \ \Rightarrow \ \ du=\dfrac{dy}{1+y^2}$, at $y=-1\to u=-\pi/4$, at $y=1\to u=\pi/4$ then
\begin{align*}
\int_{-1}^{1} \frac{e^{\tan^{-1}y}}{1+y^2}dy&=\int_{-\pi/4}^{\pi/4} e^{u}du\\
&=e^u\bigg|_{-\pi/4}^{\pi/4}\\
&=e^{\pi/4 }-e^{-\pi/4 }
\end{align*}