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.