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!
