不定积分小弟最近才学积分,有些问题不懂.在此请教几个积分的问题-.x^3+x x^3+2 2x^3-x^2 x1.∫--

1
∫(x^3+x)dx/(3x^2+2x+1)=(1/3)∫xdx-(2/9)∫dx+(5/27)∫d(3x^2+2x+1)/(3x^2+2x+1)
-(10/27)∫dx/[3(x+1/3)^2+2/3]
=(1/3)∫xdx-(2/9)∫dx+(5/27)∫d(3x^2+2x+1)/(3x^2+2x+1)
-(10/27)(1/√2)∫d(3x/√2+1/√2)/[(3x/√2+1/√2)^2+1]
=(1/3)∫xdx-(2/9)∫dx+(5/27)∫d(3x^2+2x+1)/(3x^2+2x+1)
-[10/(27√2)]arctan[(3x+1)/√2)]
=x^2/6-2x/9+(5/27)ln(3x^2+2x+1)-(10/27√2)arctan[(3x+1)/√2]+C
2
∫(x^3+2)dx/(x^2+1)^2=∫xdx/(x^2+1)-∫(x+1/2)dx/(x^2+1)^2+(5/2)∫dx/(x^2+1)^2
=(1/2)ln(x^2+1)+1/(2x^2+2)+(5/2)∫dx(x^2+1)^2
x=tanu,dx=(secu)^2,1+x^2=(secu)^2,sin2u=2x/(1+x^2)
∫dx/(x^2+1)^2=∫cosu^2du=u/2+sin2u/4=arctanx+x/(2+2x^2)
原式=(1/2)ln(x^2+1)+(1+x)/(2+2x^2)+(5/2)arctanx=C
3
∫(2x^3-x^2)dx/(x^2+1)^2=2∫xdx/(x^2+1) -∫dx/(x^2+1)-∫xdx/(x^2+1)^2-∫dx/(x^2+1)^2
=ln(x^2+1)-arctanx+1/(2x^2+2)-arctanx-x/(2+2x^2)+C
4
∫xdx/(1+x^3)=(1/3)∫[(x+1)^2-(x^2-x+1)]dx/(1+x^3)
=(1/3)∫(1+x)dx/(x^2-x+1)-(1/3)∫dx/(1+x)
=(1/6)∫(2x-1)dx/(x^2-x+1)-(1/3)∫d(x+1)/(1+x)+(1/3)∫(3/2)dx/(x^2-x+1)
=(1/6)ln(x^2-x+1)-(1/3)ln(1+x)+(1/3)(3/2)∫dx/(x^2-x+1)
=(1/6)ln(x^2-x+1)-(1/3)ln(1+x)+(1/2)∫dx/[(x-1/2)^2+3/4]
=(1/6)ln(x^2-x+1)-(1/3)ln(1+x)+(1/√3)∫d(2x/√3-1/√3)/[(2x/√3-1/√3)^2+1]
=(1/6)ln(x^2-x+1)-(1/3)ln(1+x)+(1/√3)arctan(2x/√3-1/√3)+C

难为你了，你写的题目都好变态啊。没有什么好的方法，只有用笨方法了。把分子折开，分成多项，一项项积