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 #