How do you find #int (x - 5x^2) / ((x+2)(x-1)(x-3)) dx# using partial fractions?

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Answer 1 · 2017-01-29T15:06:16

The answer is #=-22/15ln(|x+2|)+2/3ln(|x-1|)-21/5ln(|x-3|) + C#

Let's work out the decomposition into partial fractions

#(x-5x^2)/((x+2)(x-1)(x-3))#

#=A/(x+2)+B/(x-1)+C/(x-3)#

#=(A(x-1)(x-3)+B(x+2)(x-3)+C(x+2)(x-1))/((x+2)(x-1)(x-3))#

The denominators are the same, we can compare the numerators

#x-5x^2=A(x-1)(x-3)+B(x+2)(x-3)+C(x+2)(x-1)#

Let #x=-2#, #=>#, #-22=15A#, #=>#, #A=-22/15#

Let #x=1#, #=>#, #-4=-6B#, #=>#, #B=2/3#

Let #x=3#, #=>#, #-42=10C#, #=>#, #C=-21/5#

Therefore,

#(x-5x^2)/((x+2)(x-1)(x-3))=(-22/15)/(x+2)+(2/3)/(x-1)+(-21/5)/(x-3)#

So,

#int((x-5x^2)dx)/((x+2)(x-1)(x-3))=-22/15intdx/(x+2)+2/3intdx/(x-1)-21/5intdx/(x-3)#

#=-22/15ln(|x+2|)+2/3ln(|x-1|)-21/5ln(|x-3|) + C#