How do you integrate $int (x-3x^2)/((x-6)(x-2)(x-5))$ using partial fractions?

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Answer 1 · 2016-07-04T06:09:35

$int(x-3x^2)/((x-6)(x-2)(x-5))dx$

= $-121/8ln(x-6)+5/8ln(x-2)+35/4ln(x-5)+c$

Let us first convert $(x-3x^2)/((x-6)(x-2)(x-5))$ into partial fractions.

$(x-3x^2)/((x-6)(x-2)(x-5))hArrA/(x-6)+B/(x-2)+C/(x-5)$

$(x-3x^2)/((x-6)(x-2)(x-5))hArr(A(x-2)(x-5)+B(x-6)(x-5)+C(x-6)(x-2))/((x-6)(x-2)(x-5))$

= $(A(x^2-7x+10)+B(x^2-11x+30)+C(x^2-8x+12))/((x-6)(x-2)(x-5))$

= $(x^2(A+B+C)-x(7A+11B+8C)+(10A+30B+12C))/((x-6)(x-2)(x-5))$

Hence $A+B+C=-3$ , $7A+11B+8C=-1$ and $10A+30B+12C=0$

Subtracting $7$ times of first equation from second and $10$ times first from third equation, we get

$4B+C=20$ and $20B+C=30$ which gives us $B=5/8$ and $C=140/8=35/4$ and putting these in $A+B+C=-3$ , we get $A=-121/8$

Hence, $(x-3x^2)/((x-6)(x-2)(x-5))=-121/(8(x-6))+5/(8(x-2))+35/(4(x-5))$

and $int(x-3x^2)/((x-6)(x-2)(x-5))dx=int[-121/(8(x-6))+5/(8(x-2))+35/(4(x-5))]dx$

= $-121/8ln(x-6)+5/8ln(x-2)+35/4ln(x-5)+c$

How do you integrate int (x-3x^2)/((x-6)(x-2)(x-5)) int (x-3x^2)/((x-6)(x-2)(x-5)) using partial fractions?