Question

Use partial fraction: integral of (2x^2-x-20)/(x^2+x-6)

Answer 1

I got AX-2A+BX+3B = -3X -8

Answer 2

Which would mean that B=-14/5 and A= -29/5

Answer 3

I got the answer = \[2X + \frac{ -29\ln(x+3) }{ 5 }+\frac{ -14\ln(x-2) }{ 5 }\]
but the internet tells me this is the wrong answer

Answer 4

Dis you use long division first? That's not clear.

\(\dfrac{2x^{2}-x-20}{x^{2} + x - 6} = 2 - \dfrac{3x+8}{(x+3)(x-2)} = 2 - \left(\dfrac{A}{x+3} + \dfrac{B}{x-2}\right)\)

Then \(A(x-2) + B(x+3) = 3x + 8 \implies A = \dfrac{1}{5}\;and\;B = \dfrac{14}{5}\)

Well, then... It does appear that you wandered off somewhere in your algebra.

Answer 5

\[\frac{ 17 }{ 5 }\ln(x+3)-\frac{ 22 }{ 5 }\ln(x-2)\]

Answer 6

@tkhunny, I did the same thing you did except I distibuted the - sign before solving for A and B. Is this wrong?

Answer 7

That should not matter, as long as you keep track.

You just did some of your algebra wrong when solving for A and B. Do it over and be more careful. My version was easier to get right because I had fewer sign considerations.

Answer 8

So I set A(x-2) + B(X+3) = -3X-8

Answer 9

Try to be more consistent with your notation. Normally, x and X are not the same thing. I realize it is a small thing to everyone's brain, but consistency will lead to fewer errors.

Answer 10

Is my previous post right, in that I negated 3x+8, but not A+B?

Answer 11

I'm telling you it makes no difference as long as you are consistent. Your setup was fine. Your algebra wasn't. Just do it over exactly like you did in the first place. Only this time, don't make the same error.

Answer 12

I got:
\[\int\limits_{}^{}(2-\frac{ 1 }{ 5(x+3) }-\frac{ 14 }{ 5(x-2) }\]
Which should be:
\[(2x-\frac{ ln(x+3) }{ 5 }-\frac{ 14ln(x-2) }{ 5 }\]
What am I doing wrong?

Answer 13

Why do you think you're doing anything wrong?

Answer 14

The internet told me so. I switched it from 'x-2' to '2-x' and it decided I was right. This was a trick someone told me to do on the last one does anyone know why?

Answer 15

I really, REALLY hate that. We should not be spending time on formatting when we are trying to learn mathematics.

-(x-2) = 2-x

The are the same.

Answer 16

Doesn't putting a negative out fron change the answer? isn't -u different than u?

Answer 17

Do we have the COMPLETE problem statement?

Note that your integral expressions are not correct.

\(\int \dfrac{1}{x}\;dx = \ln|x| + C\)

Without additional information concerning the Domain, the absolute values are NOT optional.

Answer 18

Here are the two versions of the problem. It adds the +c at the end of the problem

Answer 19

That's just silly. The problem should be corrected.

We have ln|x+3| + ln|x-2|. These expressions must exist. Their arguments must be positive.

If we take x > 2, then x+3 > 0 and x-2 > 0 and ln(x+3) + ln(x-2) is good.
If we take -3 < x < 2, then x+3 > 0 and x-2 < 0 and ln(x+3) + ln(2-x) is good.
If we take x < -3, then x+3 < 0 and x-2 < 0 and ln(-x-3) + ln(2-x) is good.

There is NO general solution in a single expression unless we know more of the Domain. If somewhere in the problem statement, someone said -3 < x < 2, then the given answer is correct. If no one said that, it is not correct.

Answer 20

Thx