Question

求不定积分 ∫ (1→0) dx / (e^x + e^-x)

Answer 1

首瞎顷先,你要知此首道的是以磨扒陆下定积分：∫ 1 / (1+x^2) dx = arctan x + C,其中C为任意的常数.换元 e^x = t,e^x dx = dt∫ dx / (e^x + e^-x) = ∫ e^xdx / (e^2x + 1) = ∫ dt / (1+t^2) = arctan t + C = arctan (e^x) + C...