∫(2x+3)⼀(x^2+1) dx的不定积分,求过程

2024-11-08 14:10:16
推荐回答(2个)
回答1:


主要步骤如上图。

回答2:

∫[(2x-3)/(x²-2x+2)]dx
=∫(2x-2)/(x²-2x+2) dx-∫1/(x²-2x+2) dx
=∫d(x²-2x+2)/(x²-2x+2) dx-∫d(x-1)/[(x-1)²+1]
=ln(x²-2x+2)-arctan(x-1)+C