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!
