∫dx/{[(x+1)^2][(x-1)^4]}^(1/3)的不定积分积分

Question

∫dx/{[(x+1)^2][(x-1)^4]}^(1/3)的不定积分积分

Answer 1

∫dx/[(x+1)²(x-1)⁴]^(1/3)的不定积分积分
原式=∫dx/{[(x+1)^(2/3)][(x-1)^(4/3)]}
=∫dx/{[(x+1)^(2/3)][(x-1)^(-2/3)](x-1)²}
=∫{[(x-1)/(x+1)]^(2/3)}dx/(x-1)²
=-(1/2)∫{(x+1)/(x-1)]^(-2/3)}d[(x+1)/(x-1)]
=-(1/2)(3)[(x+1)/(x-1)]^(1/3)+C
=-(3/2)[(x+1)/(x-1)]^(1/3)+C