How do you integrate $int (6-15x)/( (x-1) (x+2) (x-4))$ using partial fractions?

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Answer 1 · 2016-05-04T22:52:38

$color(red)(int (6-15x)/((x-1)(x+2)(x-4)) dx=)$

$color(red)(ln(x-1)+2ln(x+2)-3ln(x-4)+C_0)$

from the given integral, set up the equation using the variables A, B, and C

$int (6-15x)/((x-1)(x+2)(x-4)) dx=int(A/(x-1)+B/(x+2)+C/(x-4))dx$

$(6-15x)/((x-1)(x+2)(x-4))=A/(x-1)+B/(x+2)+C/(x-4)$

$(6-15x)/((x-1)(x+2)(x-4))=(A(x^2-2x-8)+B(x^2-5x+4)+C(x^2+x-2))/((x-1)(x+2)(x-4))$

$(6-15x)/((x-1)(x+2)(x-4))$
$=((A+B+C)x^2+(-2A-5B+C)x^1+(-8A+4B-2C)x^0)/((x-1)(x+2)(x-4))$

The equations will be

$A+B+C=0$
$-2A-5B+C=-15$
$-8A+4B-2C=6$

Simultaneous solution results to:

$A=1$ and $B=2$ and $C=-3$

final answer

$color(red)(int (6-15x)/((x-1)(x+2)(x-4)) dx=)$

$color(red)(ln(x-1)+2ln(x+2)-3ln(x-4)+C_0)$

God bless....I hope the explanation is useful.

How do you integrate int (6-15x)/( (x-1) (x+2) (x-4))int (6-15x)/( (x-1) (x+2) (x-4)) using partial fractions?