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

1 Answer
Apr 17, 2018

#int(2x+3)/(x^2-9)dx=(lnabs(x+3)+3lnabs(x-3))/2+C#

Explanation:

Since the numerator is in a lower degree than the denominator, we can proceed to the next step.

#int(2x+3)/(x^2-9)dx# Factor

#int(2x+3)/((x+3)(x-3))dx#

We can rewrite the rational function in the form:

#A/(x+3)+B/(x-3)#

#=>A/(x+3)* (x-3)/(x-3)+B/(x-3)*(x+3)/(x+3)#

#=>(A(x-3))/((x+3)(x-3))+(B(x+3))/((x-3)(x+3))#

#=>(A(x-3)+B(x+3))/((x-3)(x+3))#

#=>(Ax-3A+Bx+3B)/((x-3)(x+3))#

#=>(Ax+Bx+3B-3A)/((x-3)(x+3))#

#=>(x(A+B)+3(B-A))/((x-3)(x+3))=(2x+3)/((x+3)(x-3))#

#=>x(A+B)+3(B-A)=2x+3#

We let:

#x(A+B)=2x#

#3(B-A)=3#

#=>A+B=2#

#=>-A+B=1# Add the equations together.

#=>2B=3#

#=>B=3/2#

#=>A=1/2#

We substitute this in the original form to get:

#int1/2*1/(x+3)+3/2*1/(x-3)dx#

#=>int1/2*1/(x+3)dx+int3/2*1/(x-3)dx#

#=>1/2int1/(x+3)dx+3/2int1/(x-3)dx#

Remember that:

#int1/(x+n)dx=lnabs(x+n)#

#=>1/2*lnabs(x+3)+3/2lnabs(x-3)#

#=>lnabs(x+3)/2+(3lnabs(x-3))/2#

#=>(lnabs(x+3)+3lnabs(x-3))/2# Do you #C# why this is incomplete?

#=>(lnabs(x+3)+3lnabs(x-3))/2+C#

That is the answer!