How do you integrate #int (x-3x^2)/((x-6)(x-2)(x+4)) # using partial fractions?

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Answer 1 · 2017-01-09T05:34:45

The answer is #=-51/20ln(∣x-6∣)+5/12ln(∣x-2∣)-13/15ln(∣x+4∣)+C#

Let's perform the decomposition into partial fractions

#(x-3x^2)/((x-6)(x-2)(x+4))=A/(x-6)+B/(x-2)+C/(x+4)#

#=(A(x-2)(x+4)+B(x-6)(x+4)+C(x-6)(x-2))/((x-6)(x-2)(x+4))#

Therefore,

#x-3x^2=A(x-2)(x+4)+B(x-6)(x+4)+C(x-6)(x-2)#

Let #x=6#, #=>#, #-102=40A#, #=>#, #A=-51/20#

Let #x=2#, #=>#, #-10=-24B#, #=>#, #B=5/12#

Let #x=-4#, #=>#, #-52=60C#, #=>#, #C=-13/15#

So,

#(x-3x^2)/((x-6)(x-2)(x+4))=(-51/20)/(x-6)+(5/12)/(x-2)+(-13/15)/(x+4)#

Therefore,

#int((x-3x^2)dx)/((x-6)(x-2)(x+4))=-51/20intdx/(x-6)+5/12intdx/(x-2)-13/15intdx/(x+4)#

#=-51/20ln(∣x-6∣)+5/12ln(∣x-2∣)-13/15ln(∣x+4∣)+C#