✔ 最佳答案
1 ∫(x^4+3x^3+2x^2+x-4)/(x+1)^5dx
Sol
Set (x^4+3x^3+2x^2+x-4)/(x+1)^5
=a/(x+1)+b/(x+1)^2+c/(x+1)^3+d/(x+1)^4+e/(x+1)^5
(x^4+3x^3+2x^2+x-4)=a(x+1)^4+b(x+1)^3+c(x+1)^2+d(x+1)+e
1 3 2 1 -4|
-1 -2 0 -1|-1
──────────────
1 2 0 1|-5=e
-1 -1 1|-1
───────────
1 1 -1| 2=d
-1 0|-1
────────
1 0|-1=c
-1|-1
─────
1|-1=b
a=1
So
(x^4+3x^3+2x^2+x-4)=1(x+1)^4-1/(x+1)^3-1(x+1)^2+2(x+1)-5
∫(x^4+3x^3+2x^2+x-4)/(x+1)^5dx
=∫1/(x+1)dx-∫1/(x+1)^2dx-∫1/(x+1)^3dx+2∫1/(x+1)^4dx-5∫1/(x+1)^5dx
=ln|x-1|+1/(x-1)+1/2(x-1)^2-2/3(x-1)^3+5/4(x-1)^4+c 2 ∫(4x+16)/[(x+1)^2(x+5)]dx
Sol
Set (4x+16)/[(x+1)^2(x+5)]=a/(x+1)+b/(x+1)^2+c/(x+5)
4x+16=a(x+1)(x+5)+b(x+5)+c(x+1)^2
when x=-1
12=b(4)
b=3
when x=-5
-4=c*16
c=-1/4
4x+16=a(x+1)(x+5)+3(x+5)-(1/4)(x+1)^2
16x+64=4a(x^2+6x+5)+12x+60-x^2-2x-1
4a(x^2+6x+5)=x^2+6x+5
a=1/4
(4x+16)/[(x+1)^2(x+5)]=(1/4)/(x+1)+3/(x+1)^2-(1/4)/(x+5)
∫(4x+16)/[(x+1)^2(x+5)]dx
=(1/4)∫1/(x+1)dx+3∫1/(x+1)^2dx-(1/4)∫1/(x+5)dx
=(1/4)ln|x+1|-3/(x+1)-(1/4)ln|x+5|+c