高等数学求不定积分：这题用倒代换怎么做？

Answer 1

令 x = 1/u, 则 dx = -du/u^2
I = ∫dx/[x(x^3+1)] = ∫-du/[u^2(1/u)(1/u^3+1)]
= ∫-du/(1/u^2+u) = ∫-u^2du/(1+u^3)
= -(1/3)ln|1+u^3|+C = -(1/3)ln|1+1/x^3| +C
= -(1/3)ln|(x^3+1)/x^3| +C = (1/3)ln|(x^3/(1+x^3)| +C