How do you use partial fraction decomposition to decompose the fraction to integrate $(x^2-x-8)/((x+1)(x^2+5x+6))$ ?

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Answer 1 · 2015-08-13T18:49:29

$(x^2-x-8)/((x+1)(x+2)(x+3))= -3/(x+1) + 2/(x+2) + 2/(x+3)$

$(x^2-x-8)/((x+1)(x^2+5x+6))$

First we need to finish factoring the denominator:

$(x^2-x-8)/((x+1)(x+2)(x+3))$

Now we want $A, B, "and " C$ to get:

$A/(x+1) + B/(x+2) + C/(x+3) = (x^2-x-8)/((x+1)(x+2)(x+3))$

So we need:

$A(x+2)(x+3)+B(x+1)(x+3)+C(x+1)(x+2) = x^2-x-8$

The left side can be rewritten:

$Ax^2+5Ax+6A+Bx^2+4Bx+3B+Cx^2+3Cx+2C$

and then:

$(A+B+C)x^2 +(5A+4B+3C)x+(6A+3B+2C)=x^2-x-8$

So we need to solve the system:

$A+B+C=1$

$5A+4B+3C=-1$

$6A+3B+2C=-8$

Multiplying the first equation by $-3$ and adding to the second. Then the first times $-2$ and add to the third, gets us:

$2A+B=-4$

$4A+B=-10$

So $2A = -6$ and $A = -3$ , which gets us $B = 2$ and together these give us $C=2$ .

$(x^2-x-8)/((x+1)(x+2)(x+3))= -3/(x+1) + 2/(x+2) + 2/(x+3)$

How do you use partial fraction decomposition to decompose the fraction to integrate (x^2-x-8)/((x+1)(x^2+5x+6))(x^2-x-8)/((x+1)(x^2+5x+6))?