Given
$$
\int \exp(x) \cosh(x) dx
$$
Write this as
$$
\int \exp(x) \cosh(ax) dx
$$
then
$$
\begin{eqnarray}
\int \exp(x) \cosh(ax) dx &=& \exp(x) \cosh(ax) - a \int \exp(x) \sinh(ax) dx\\
&=& \exp(x) \cosh(ax) - a \exp(x) \sinh(ax) + a^2 \int \exp(x) \cosh(ax) dx
\end{eqnarray}
$$
thus
$$
\begin{eqnarray}
\int \exp(x) \cosh(ax) dx &=& \frac{ \exp(x) \cosh(ax) - a \exp(x) \sinh(ax)}
{ (1 - a)(1+a) }
\end{eqnarray}
$$
whence
$$
\begin{eqnarray}
\int \exp(x) \cosh(x) dx &=& \lim_{a \rightarrow 1} \frac{ \exp(x) \cosh(ax) - a \exp(x) \sinh(ax)}{ (1 - a) (1 + a) }
\end{eqnarray}
$$
The last part - taking the limit gives
$$
\begin{eqnarray}
\int \exp(x) \cosh(x) dx &=& \lim_{a \rightarrow 1} \frac{ \exp(x) \cosh(ax) - a \exp(x) \sinh(ax)}{ (1 - a) (1 + a) }\\
&=& \lim_{a \rightarrow 1} \frac{x \exp(x) [ \sinh(ax) - a \cosh(ax)] - \exp(x) \sinh(ax)}{-2}\\
&=& \frac{1}{2} \exp(x)\sinh(x) + \frac{1}{2} x
\end{eqnarray}
$$