Answer
$$40$$
Work Step by Step
Given
$$\int_{-10}^{10} \frac{2 e^{x}}{\sinh x+\cosh x} d x$$
Since
\begin{aligned}
\sinh(x)&= \frac{e^x-e^{-x}}{2}\\
\cosh(x)&= \frac{e^x+e^{-x}}{2}\\
\end{aligned}
Then
\begin{aligned}
\sinh(x) + \cosh(x) &= \frac{e^x-e^{-x}}{2}+ \frac{e^x+e^{-x}}{2}\\
& = \frac{2e^x}{2}\\
&=e^x
\end{aligned}
Hence
\begin{aligned}
\int_{-10}^{10} \frac{2 e^{x}}{\sinh x+\cosh x} d x&= \int_{-10}^{10} \frac{2 e^{x}}{e^ x} d x\\
&=2\int_{-10}^{10}dx\\
&=2(x)\bigg|_{-10}^{10}\\
&= 2(10-(-10))\\
&= 40
\end{aligned}