Question

求不定积分∫[3(x^2)+2x]/(x+1) dx

Answer 1

解：
∫(3x²+2x)/(x+1) dx
=3∫(x²+2x/3)/(x+1)dx
=3∫(x²+2x+1-4x/3-1)/(x+1) dx
=3∫[(x+1)-(4x/3+4/3-1/3)/(x+1)]dx
=3∫[(x+1)-4/3+1/3·1/(x+1)]dx
=3∫(x+1) d(x+1)-4∫dx+∫1/(x+1)dx
=3(x+1)²/2-4x+ln|x+1|+C

追问

=3∫[(x+1)-(4x/3+4/3-1/3)/(x+1)]dx???

追答

=3∫(x²+2x+1-4x/3-1)/(x+1) dx
=3∫[(x+1)-(4x/3+1)/(x+1)]dx

=3∫[(x+1)-(4x/3+4/3-1/3)/(x+1)]dx

Answer 2

3(x^2)+2x=3(x+1)^2 - 4(x+1) + 1

[3(x^2)+2x]/(x+1)=3(x+1) - 4 + 1/(x+1)

然后再分别积分，再加起来，over