How do you find #int ( x + 1)/(x^3 + x^2 -2x) dx# using partial fractions?

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Answer 1 · 2018-03-13T10:17:37

The answer is #=-1/2ln(|x|)-1/6ln(|x+2|)+2/3ln(|x-1|)+C#

Perform the decomposition into partial fractions

#(x+1)/(x^3+x^2-2x)=(x+1)/(x(x^2+x-2))#

#=(x+1)/(x(x+2)(x-1))#

#=A/(x)+B/(x+2)+C/(x-1)#

#=(A(x+2)(x-1)+B(x)(x-1)+C(x)(x+2))/(x(x+2)(x-1))#

The denominators are the same, compare the numerators

#x+1=A(x+2)(x-1)+B(x)(x-1)+C(x)(x+2)#

Let #x=0#, #=>#, #1=-2A#, #=>#, #A=-1/2#

Let #x=-2#, #=>#, #-1=6B#, #=>#, #B=-1/6#

Let #x=1#, #=>#, #2=3C#, #=>#, #C=2/3#

Therefore,

#(x+1)/(x^3+x^2-2x)=(-1/2)/(x)+(-1/6)/(x+2)+(2/3)/(x-1)#

#int((x+1)dx)/(x^3+x^2-2x)=int(-1/2dx)/(x)+int(-1/6dx)/(x+2)+int(2/3dx)/(x-1)#

#=-1/2ln(|x|)-1/6ln(|x+2|)+2/3ln(|x-1|)+C#