Calculators TI-89 Titanium INTEGRATION Flaw?

1. Feb 22, 2007

Switch_fx

TI-89 Titanium INTEGRATION Flaw???

Ok so this is my first post, and I am in dire need of assistance. I have a big exam tomorrow and if I can't figure out why my TI-89 is screwing up this simple integration I am in some trouble...

I am looking for the integral of : 1/[G-(Cv)/M] with respect to "v"

(erroneus, G=gravitational constant, C=drag coefficient, M=mass, and V=velocity)

anyways...
I know the correct answer is : -(M/C)[ln(G-(Cv)/M)]

but my TI-89 calculator gives back the answer as:

-(M/C)[ln(|Cv-GM|)]

If anyone could please explain to me why this is so or what I am doing wrong it would be of great help...and yes I am inputting the integral in the correct fashion

Thanks to anyone who can help!

~SWITCH

2. Feb 22, 2007

d_leet

If you get rid of the fraction in the denominator in the original function you will arrive at the answe your calculator does, however you should notice that these only differ by a constant and that when you integrate you always have to add on a constant of integration.

3. Feb 22, 2007

Switch_fx

Why would I get rid of the fraction in the denominator? And if I get rid of it what do I replace it with?

4. Feb 22, 2007

HallsofIvy

Both are missing the added constant. d_leet's point is that your and the calculator's answer differ only by a constant.

5. Feb 22, 2007

ssd

If we leave the constant part, the calculator is right in general, as it gives the absolute value of quantity inside logarithm.

Last edited: Feb 22, 2007
6. Feb 22, 2007

DrKareem

Maybe if it's hard for you to integrate 1/(1-ax) with respect to x then you do have a problem?

Taken from 1/a onwards, the integral of 1/1-ax is (-1/a) ln |1-ax|. Knowing how that solve your problem without needing the calculator.

7. Feb 23, 2007

arunma

Take the "correct" answer, and use the logarithmic rule for division: ln(a/b) = ln(a) - ln(b). Normally I'd get in trouble for telling you that, but you could have just as easily looked it up in the textbook, so I thought I'd save you a couple minutes of tedious page turning. Anyway, considering what the others here have told you, just by expanding your answer in this way, you should be convinced that your answer and the 89's answer are the same.