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,

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

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

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