When you are at this point:
$\displaystyle \int e^{2x-1}\,dx=\sum_{n=0}^{\infty}\frac{(2x-1)^{n+1}}{2(n+1)!}$
You might then observe that we can write:
$\displaystyle \int e^{2x-1}\,dx=\frac{1}{2}\left(\sum_{n=1}^{\infty}\frac{( 2x-1)^{n}}{n!}\right)=\frac{1}{2}\left(e^{2x-1}-1\right)$
As an addendum, we should probably write:
$\displaystyle \int e^{2x-1}\,dx=\sum_{n=0}^{\infty}\frac{(2x-1)^{n+1}}{2(n+1)!}+C$
And then we'd end up with:
$\displaystyle \int e^{2x-1}\,dx=\frac{1}{2}\left(e^{2x-1}-1\right)+C$
And it we define:
$\displaystyle C_1=C-\frac{1}{2}$
We could write:
$\displaystyle \int e^{2x-1}\,dx=\frac{1}{2}e^{2x-1}+C_1$