Question

x³/x+1不定积分？？？请写详细点，谢谢

Answer 1

以上，请采纳。

Answer 2

∫[x^3/(x+1)]dx = ∫{[(x^3+x^2)-(x^2+x)+(x+1)-1]/(x+1)}dx
= ∫[x^2 - x + 1 - 1/(x+1)]dx
= x^3/3 - x^2/2 + x - ln|x+1| + C

Answer 3

楼上小日本阵垃圾

追问

请问你会吗，我没怎么看懂

追答

哪里不懂

追问

分子怎么变的？

追答

追问

懂了谢谢谢

追答

望采纳

追问

我为什么采纳不了，刷新了好几次了

追答

已经采纳了

Answer 4

∫1/(x³+2x²+2x+1)dx
=∫1/[(x+1)(x²+x+1)]dx
=∫[(x²+x+1)-(x²+x)]/[(x+1)(x²+x+1)]dx
=∫[1/(x+1)-x/(x²+x+1)]dx
=ln|x+1|-∫x/(x²+x+1)dx
=ln|x+1|-1/2·∫(2x+1-1)/(x²+x+1)dx
=ln|x+1|-1/2·∫(2x+1)/(x²+x+1)dx+1/2·∫1/(x²+x+1)dx
=ln|x+1|-1/2·ln(x²+x+1)+1/2·∫1/[(x+1/2)²+3/4]dx
=ln|x+1|-1/2·ln(x²+x+1)+1/2·1/√(3/4)·arctan[(x+1/2)/√(3/4)]+C
=ln|x+1|-1/2·ln(x²+x+1)+1/√3·arctan[(2x+1)/√3]+C