# [SOLVED] Definite Integral, &amp; Partial fractions

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• Sep 15th 2008, 09:59 PM
matt3D
[SOLVED] Definite Integral, &amp; Partial fractions
Hi, I have the problem: $\int_{-1}^{3} \frac {x^3+3}{x^2+10x+21}dx$ The denominator was actually $(x+7)(x+3)$ but I know that the degree is higher in the numerator so polynomial long division needs to occur. After I do polynomial division I get $x^2+10x+21 \times \frac {x-10}{207+79x}$after this I get stuck. Usually the denominator has a $x^2$ or higher, then I could break it up but this time it doesn't.
Thanks,
Matt
• Sep 15th 2008, 10:04 PM
Jameson
I'll just say that I think this should break up like you're talking about. The degree of the numerator is larger than the denominator so you will have a remainder unless it divides evenly, which it doesn't. Try the division again. I'll check it soon.
• Sep 15th 2008, 10:09 PM
Jameson
Ok I did it using online software to be fast :) but i get $x-10+\frac{79x+210}{x^2+10x+21}$.
• Sep 15th 2008, 10:15 PM
matt3D
oops, I actually meant $x-10+ \frac {213+79x}{x^2+10x+21}$ I got 213 not 210...? in the long division at the end I have 3-(-210)...
• Sep 15th 2008, 10:16 PM
Jameson
Quote:

Originally Posted by matt3D
oops, I actually meant $x^2+10x+21+ \frac {207+79x}{x^2+10x+21}$

The first term shouldn't be the denominator of the original term, it should be the answer to the division and the second term is the remainder. Make sense?
• Sep 15th 2008, 10:20 PM
matt3D
Quote:

Originally Posted by Jameson
The first term shouldn't be the denominator of the original term, it should be the answer to the division and the second term is the remainder. Make sense?

Yea, I got it now... but our answers for the long division are different.
• Sep 15th 2008, 10:22 PM
Jameson
Quote:

Originally Posted by matt3D
Yea, I got it now... but our answers for the long division are different.

Go to this site, long divide polynomials, type in your two poly. and it will divide them and show the work.
• Sep 15th 2008, 10:22 PM
11rdc11
Quote:

Originally Posted by matt3D
Yea, I got it now... but our answers for the long division are different.

Show us your steps for the long divison
• Sep 15th 2008, 10:32 PM
matt3D
My answer is correct, I just confirmed it by that website...(Nod)
• Sep 15th 2008, 10:33 PM
Jameson
Quote:

Originally Posted by matt3D
My answer is correct, I just confirmed it by that website...(Nod)

Great! Sorry about my error then. Can you finish the problem now?
• Sep 15th 2008, 10:34 PM
matt3D
Quote:

Originally Posted by Jameson
Great! Sorry about my error then. Can you finish the problem now?

No, heh, after that I need help... (Wait)
• Sep 15th 2008, 10:38 PM
11rdc11
Quote:

Originally Posted by matt3D
No, heh, after that I need help... (Wait)

Nope jameson is right. Multiply your answer out and you will see you don't get the original equation
• Sep 15th 2008, 10:43 PM
Jameson
Ok so if this is the correct division then let's look at it.

The first term is easy obviously.

The second looks like it could be a u-sub problem, but it won't quite work so like your title says, let's use partial fractions.

Factoring the denom. we get (x+3) and (x+7). So we know that this should expand into two terms with some constant term in the numerator. Let's say that $\frac{79x+210}{x^2+10x+21}=\frac{A}{x+3}+\frac{B}{ x+7}$. Now to solve for A and B multiply all terms by the denom. of the first term and you'll get $79x+210=A(x+7)+B(x+3)$. The easiest way to find A and B now is first let x=-7 and find B, then x=-3 and find A.

With me?
• Sep 15th 2008, 10:44 PM
Jameson
Quote:

Originally Posted by 11rdc11
Nope jameson is right. Multiply your answer out and you will see you don't get the original equation

Cool I was right. Above post edited for this now.
• Sep 15th 2008, 10:50 PM
11rdc11
Yep the first term after doing long division can't be $x^2$. It would have to be x because x times $x^2$ equals $x^3$. Now if the first term was $x^2$ we would have $x^2$ times $x^2$ which would be $x^4$

Oops stand corrected! I didn't notice matt3d corrected his long division and he is correct it 213 not 210. Well everything Jameson said to do after with the partial fractions is correct. We all made minor mistakes lol.
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