求x/(e^x+e^(1-x))dx的定积分(x在0到1) 要有过程

Question

求x/(e^x+e^(1-x))dx的定积分(x在0到1)　要有过程

Answer 1

令u=1-x，du=-dx
J = ∫[0，1] x/[e^x+e^(1-x)] dx
= -∫[1，0] (1-u)/[e^u+e^(1-u)] du
= ∫[0，1] 1/[e^u+e^(1-u)] du - ∫[0，1] u/[e^u+e^(1-u)] du
= K - J
2J = K => J=(1/2)K
J = (1/2)∫[0，1] 1/(e^u+e/e^u) du，t=e^u，u=lnt，du=1/t dt
= (1/2)∫[1，e] 1/[(t+e/t)t] dt
= (1/2)∫[1，e] 1/(t²+e) dt
= (1/2) * 1/√e * arctan(t/√e)，[1，e]，公式∫ dx/(a²+x²) = (1/a)arctan(x/a)
= 1/(2√e) * [arctan(√e) - arctan(1/√e)]
= 1/(2√e) * arctan[(√e-1/√e)/(1+√e*1/√e)]，公式arctanx-arctany=arctan[(x-y)/(1+xy)]，xy>-1
= 1/(2√e) * arctan[(e-1)/√e * 1/2]
= 1/(2√e) * arctan[(e-1)/(2√e)]本回答被提问者采纳