AP Calculus help, Evaluate the improper integral or state that it diverges.?

Question

Evaluate the improper integral or state that it diverges.

dx/(x^2-9) from -4 to negative infinity
(-4 on top, negative infinity on bottom).

I've done most of it so far, I think, and this is what I have.
lim a->neg. infinty from -b to a (-b on top, a on bottom) (1/((x-3)(x+3)).
So that equals (a/(x-3)) + (b/(x+3)).

I set up the equation 1=ax + 3a + bx - 3b.
So 3a - 3b = 1 and a + b = 0.
Therefore A = (1/6) and B = (-1/6).

And that is where I am stuck, what do I do next?

The possible answers are:
A) (1/3)ln(7)
B) (1/3)ln(1/7)
C) (1/6)ln(7)
D) (-1/6)ln(7)

Any help will be appreciated, thanks a lot!

Answer 1

You are right to start with partial fractions.

1/(x² - 9) = (1/6)/(x - 3) - (1/6)/(x + 3)

Now you can integrate.

∫{1/(x² - 9)}dx = (1/6)∫{1/(x - 3) - 1/(x + 3)}dx
= (1/6)(ln(x - 3) = ln(x + 3)}
= (1/6)ln[(x - 3) / (x + 3)] | {evaluated on (-∞, -4]}
= (1/6)[(ln 7) - ln(1)] = (1/6)(ln 7)

The answer is C) (1/6)ln(7).

Answer 2

A = 1/6, B = -1/6 okay

so, ∫A/(x-3) + B/(x+3) dx
= A ln (x-3) + B ln (x+3)

= (1/6) ( ln(x-3) - ln(x+3) )

= ln( (x-3)/(x+3) ) on [-∞.-4]


= ln(7) / 6