Question

计算积分∫[-∞,+∞]1/[(1+x2)(1+ex)]dx

Answer 1

I=∫[-∞,+∞]1/(1+x^2)(1+e^x)dx=∫[-∝,+∝]dx/(1+x^2)(1+e^(-x))

2I=∫[-∝,+∝] [1/(1+e^x)+1/(1+e^(-x))]dx/(1+x^2)

=∫[-∝,+∝]dx/(1+x^2)

=arctanx|[-∝,+∝]

=π

I=π/2