integrate e^x coshx dx
$\displaystyle Cosh(x)=\frac{e^x+e^{-x}}{2}$
$\displaystyle e^xCosh(x)=\frac{e^xe^x+e^xe^{-x}}{2}=\frac{e^{2x}+e^0}{2}$
$\displaystyle =\frac{e^{2x}+1}{2}$
$\displaystyle \frac{1}{2}\int{e^{2x}}dx+\int{\frac{1}{2}}dx=\fra c{1}{2}\int{e^xe^x}dx+\frac{x}{2}+C$
To integrate by parts
Let $\displaystyle \int{e^xe^x}dx=I$
$\displaystyle u=e^x\ \Rightarrow\ du=e^xdx$
$\displaystyle dv=e^xdx\ \Rightarrow\ v=e^x$
$\displaystyle uv-\int{v}du=e^xe^x-\int{e^xe^x}dx$
$\displaystyle I=e^{2x}-I$
$\displaystyle 2I=e^{2x}$
$\displaystyle I=\frac{e^{2x}}{2}+C$
Therefore
$\displaystyle \int{e^xCosh(x)}dx=\frac{1}{4}e^{2x}+\frac{1}{2}x+ C$