∫x/(x+2)(x+3)²dx

∫x/(x+2)(x+3)²dx
∫x/(x+2)(x+3)²dx

∫x/(x+2)(x+3)²dx
把下边有理化,就非常容易了.

∫x/[(x+2)(x+3)²]dx
x/[(x+2)(x+3)^2]=A/(x+2)+(Bx+C)/(x+3)²
=[A(x+3)²+(Bx+C)(x+2)]/[(x+2)(x+3)²]
A(x+3)²+(Bx+C)(x+2)=x
(A+B)x²+(6A+2B+C)x+(2C+9)=x
A+B=0...

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∫x/[(x+2)(x+3)²]dx
x/[(x+2)(x+3)^2]=A/(x+2)+(Bx+C)/(x+3)²
=[A(x+3)²+(Bx+C)(x+2)]/[(x+2)(x+3)²]
A(x+3)²+(Bx+C)(x+2)=x
(A+B)x²+(6A+2B+C)x+(2C+9)=x
A+B=0
6A+2B+C=1
2C+9=0
A=11/8
B=-11/8
C=-9/2
x/[(x+2)(x+3)^2]=(11/8)/(x+2)+(-11x/8-9/2)/(x+3)²
=(1/8)[11/(x+2)-(11x+33)/(x+3)²-3/(x+3)²]
∫x/[(x+2)(x+3)²]dx
=(1/8)[∫11/(x+2)dx-∫(11x+33)/(x+3)²dx-∫3/(x+3)²dx]
=(1/8)[∫11/(x+2)dx-∫(11/2)/(x+3)²d(x+3)²-∫3/(x+3)²d(x+3)]
=(11/8)ln(x+2)-(11/16)ln(x+3)²+33/[8(x+3)]+C
=11ln[(x+2)/(x+3)]/8+33/[8(x+3)]+C
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