Evaluate ∫01exe2x+1dx\int_{0}^{1} \frac{e^x}{e^{2x}+1} dx∫01e2x+1exdx. Use substitution u=exu = e^xu=ex.
arctan(e)−π/4\arctan(e) - \pi/4arctan(e)−π/4
ln(e+1)\ln(e+1)ln(e+1)
e−1e - 1e−1
π/4\pi/4π/4