How do you integrate $int (3x-2x^2)/((x+9)(x+2)(x-5))$ using partial fractions?

Question

1451 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-04T01:34:08

$1/14(4ln(x+2)-27ln(x+9)-5ln(x-5))+C$

Given: $int(3x-2x^2)/((x+9)(x+2)(x-5))dx$ using partial fractions

Partial Fractions:
$(3x-2x^2)/((x+9)(x+2)(x-5)) = A/(x+9) + B/(x+2) + C/(x-5)$

$3x-2x^2= A(x+2)(x-5) + B(x+9)(x-5) + C(x+9)(x+2)$

$= A(x^2-3x-10)+B(x^2+4x-45)+C(x^2+11x+18)$

rearrange:

$3x-2x^2= x^2(A+B+C) +x(-3A+4B+11C) + (-10A-45B+18C)$

Write 3 equations:
(1): $" "A+B+C = -2$

(2): $" "-3A+4B+11C = 3$

(3): $" "-10A-45B+18C = 0$

Combine equations to solve:
3*(1)+(2): $3A + 3B + 3C = -6$
$" "ul(-3A+4B+11C = 3" ")$
$" "7B +14C = -3$

10(2) - 3(3): $-30A +40B+110C =30$
$" "ul(30A +135B-54C = 0" ")$
$" "175B+56C = 30$

Combine two new equations to solve for B:
$175B+56C = 30$
$ul(-28B-56C=12)$
$147B" "=42; " " B = 2/7 = 4/14$

Back-substitute to find C:
$7*2/7 +14C = -3; " "14C = -5; " " C = -5/14$

Back-substitute to find A:
$A +2/7 -5/14 =-2; " " A = -27/14$

$int(3x-2x^2)/((x+9)(x+2)(x-5))dx = int(-27 dx)/(14(x+9)) +int(2 dx)/(7(x+2)) +int(-5 dx)/(14(x-5))$

$= 1/14(4ln(x+2)-27ln(x+9)-5ln(x-5))+C$

How do you integrate int (3x-2x^2)/((x+9)(x+2)(x-5)) int (3x-2x^2)/((x+9)(x+2)(x-5)) using partial fractions?