求有理函数积分：①∫x^4/(x-1)^3dx②∫1/x^2(1+2x)dx③∫x/x^3-x^2+x-1dx.

Question

求有理函数积分：①∫x^4/(x-1)^3dx ②∫1/x^2(1+2x)dx ③∫x/x^3-x^2+x-1dx.

Answer 1

∫x^4dx/(x-1)^3
= -(1/2)∫x^4d(x-1)^(-2)
=(-1/2)x^4/(x-1)^2+(1/2)∫4x^3dx/(x-1)^2
=(-1/2)x^4/(x-1)^2+(-2)∫x^3d(x-1)^(-1)
=(-1/2)x^4/(x-1)^2-2x^3/(x-1)+2∫3x^2dx/(x-1)
=(-1/2)x^4/(x-1)^2-2x^3/(x-1) +6∫(x+1)dx+6∫dx/(x-1)
=(-1/2)x^4/(x-1)^2-2x^3/(x-1)+3(x+1)^2+6ln|x-1|+C
∫dx/[x^2(1+2x]
=∫[(1+2x)-2x]dx/[x^2(1+2x)]
=∫dx/x^2 -∫2dx/[x(1+2x)]
= -1/x-2∫[(1+2x)-2x]dx/[x(1+2x)]
=-1/x-2∫dx/x+4∫dx/(1+2x)
= -1/x-2ln|x|+2ln|1+2x|+C
∫xdx/(x^3-x^2+x-1)
=∫xdx/[(x^2+1)(x-1)]
=(1/2)∫[(x^2+1)-(x-1)^2]dx/[(x^2+1)(x-1)]
=(1/2)∫dx/(x-1)-(1/2)∫(x-1)dx/[x^2+1)
=(1/2)ln|x-1|-(1/4)ln(x^2+1)+(1/2)arctanx+C