Expert: Ahmed Salami - 2/28/2010

Question
I'm copying this from Yahoo!, so please excuse my colloquial descriptions; they were directed towards my peers:

Please help me figure out where I screwed up? It's #17 of section 8.4 in James Stewart's Calculus 6e. It's "Integration of Rational Functions by Partial Fractions". Here's the problem--Oh, and this is how I'll write integrals: S[a,b] f(x)dx , where [a,b] are the lower and upper limits of integration. Integrated but unsolved integrals will be expressed [f(x) | [a,b]. Okay, here we go. PLEASE READ THE DENOMINATOR CAREFULLY!:

S[1,2] (4y^2 - 7y - 12)dy/y(y + 2)(y - 3) Okay, got it? Nothing tricky about the dy. 3 terms in the denominator. After looking at it, I wrote it as a series of Fractions in terms of A, B, and C:

(4y^2 - 7y - 12)/y(y + 2)(y - 3) = A/y + B/(y + 2) + C/(y - 3) From there I multiplied both sides by the denominator on the left. Assuming I made no mistakes, this got me:

4y^2 - 7y - 12 = A(y + 2)(y - 3) + By(y - 3) + Cy(y + 2) I don't think that's wrong. Then I grouped A, B, and C in terms of y:

4y^2 - 7y - 12 = (A + B + C)y^2 + (-A - 3B + 2C)y - 6A Then I wrote a system of 3 equations:

1) A + B + C = 4
2) -A - 3B + 2C = -7
3) -6A = -12

Solving the system got me A = 2, B = 9/5, and C = 1/5. Rewriting the integral:

S[1,2](4y^2 - 7y - 12)dy/y(y + 2)(y - 3) = 2*Sdy/y + (9/5)*Sdy/(y + 2) + (1/5)*Sdy/(y - 3)

By this point I must be wrong. The third term is unsolvable (as you will see). Continuing anyway, w/ limits of integration from 1 to 2, watch that ugly third term:

= 2*[lny|[1,2] + (9/5)*[ln(y + 2)|[1,2] + (1/5)*[ln(y - 3)|[1,2]

I'm not going any farther. I was on the threshold of success and it's now 4 hours later. I can't solve a logarithm with a negative argument, so I must have screwed up. The answer in the back of the book is (including parenthetical note):

(27/5)*ln2 - (9/5)*ln3 (or (9/5)*ln(8/3))

PLEASE help me figure out what I did wrong?

Answer
Hi Spencer,
Very nice of you to show me the work that's already done, makes everything easier.
Now, you havent made any mistake except for thinking that [ln(y - 3)|[1,2] cannot be evaluated. Sure, solving a logarithm with a negative argument can be a problem but note here that there are two negative arguments involved.
[ln(y - 3)|[1,2] = ln(2-3) - ln(1-3)
                = ln[(2-3)/(1-3)]
                = ln[(-1)/(-2)]
                = ln(1/2)  OR -ln(2)   [ from ln(1/2) = ln(2^-1) = -ln(2) ]
Thats out of the way now, i'm sure you can work out the rest to the final answer. You can always get back to me.

Regards