Question

∫（sinx+cosx）/三次根号下sinx-cosx dx

Answer 1

设t=³√(sinx-cosx)
sinx-cosx=t³
(sinx+cosx)dx=3t²dt
代入易得结果为3/2t²+c
回代即可得解

Answer 2

∫ (sinx+cosx)/(sinx-cosx)^(1/3) dx
= ∫d(sinx-cosx)/(sinx-cosx)^(1/3)
= (sinx-cosx)^(-1/3+1) / (-1/3+1) + C
= (3/2)(sinx-cosx)^(2/3) + C