∫(2x+3)/(x^2+1) dx的不定积分，求过程

Answer 1

主要步骤如上图。

追答

Answer 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