Let $f(x)=\frac{e^{2x-1}}{1+e^{2x-1}}$ Then find $\sum_{k=1}^{1233}f\left(\frac{k}{1234}\right)$
How I proceed: $\sum_{k=1}^{1233}f\left(\frac{k}{1234}\right)=\int_{1}^{1233}\frac{e^{\frac{2x}{1234}-1}}{1+e^{\frac{2x}{1234}-1}}~dx$ then how I solve this integral. please help.