Question

不定积分第五题？

Answer 1

Answer 2

（5）∫dx/[(x-1)(x+1)^2]^(1/3)

=∫[(x+1)/(x-1)]^(1/3)*[1/(x+1)]dx
令t=[(x+1)/(x-1)]^(1/3)，则x=2/(t^3-1)+1，dx=-(6t^2)/(t^3-1)^2dt
原式=∫t*[(t^3-1)/(2t^3)]*[-(6t^2)/(t^3-1)^2]dt
=-3*∫dt/(t^3-1)
=-3*∫dt/(t-1)(t^2+t+1)
=-3*∫[1/(t-1)+t/(t^2+t+1)]dt
=-3*∫{1/(t-1)+(1/2)*(2t+1)/(t^2+t+1)-(1/2)/[(t+1/2)^2+3/4]}dt
=-3*ln|t-1|-(3/2)*ln|t^2+t+1|+√3*arctan[(2t+1)/√3]+C
=√3*arctan{{2*[(x+1)/(x-1)]^(1/3)+1}/√3}-3*ln|[(x+1)/(x-1)]^(1/3)-1|-(3/2)*ln{[(x+1)/(x-1)]^(2/3)+[(x+1)/(x-1)]^(1/3)+1}+C，其中C是任意常数