Question

计算不定积分∫x²3√1-xdx,麻烦写下具体过程,

Answer 1

令3√1-x=t

1-x=t³

x=1-t³

dx=-3t²dt

原式=∫(1-t³)²t(-3t²)dt

=-3∫（t^6-2t³+1）t³dt

=-3∫(t^9-2t^6+t³)dt

=-3(t^10/10-2t^7/7+t^4/4)+c

=-3/10 (3√1-x)^10+6/7 (3√1-x)^7 -3/4 (3√1-x)^4+c