How do you integrate $int (5x+12)/(x^2+5x+6)$ using partial fractions?

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Answer 1 · 2016-02-07T12:51:59

$ln(x+3)+ln(x+2) +C$

To begin we must take:

$(5x+12)/(x^2+5x+6)$ and decompose it into its partial fractions. First factorise the denominator and then split the fraction up as follows:

$(5x+12)/((x+3)(x+2)) = A/(x+3) +B/(x+2)$

Now, if we multiply the whole thing through by $(x+3)(x+2)$ then we should get an equation that will allow us to solve for $A$ and $B$ .

$-> 5x+12 = A(x+2)+B(x+3)$

Now, to find $A$ set $x=3$ to cancel the second term and we get:

$5(-3)+12 = A(-3+2)+B(-3+3)$
$-3=-A -> A = 3$

Now set $x=-2$ to obtain the value for for $B$ .

$->5(-2) +12 = A(-2+2)+B(-2+3) -> B = 2$

So now we have that $A=3$ and $B=2$ we can re write the fraction given in the question as:

$(5x+12)/(x^2+5x+6) = 3/(x+3) + 2/(x+2)$

So we can now integrate:

$int3/(x+3) + 2/(x+2) dx = 3ln(x+3)+2ln(x+2) +C$

How do you integrate int (5x+12)/(x^2+5x+6)int (5x+12)/(x^2+5x+6) using partial fractions?