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Post by Sleepy on Dec 3, 2011 13:23:33 GMT -5
Alrighty, I've got a math problem involving integrals that I'm trying to work out, and I can't seem to get it. I believe the method for solving this one is partial fraction decomposition. Here's what I start with: I think the denominator ends up being a distinct, irreducible quadratic, but I'm not quite sure. Any math geeks that can help me solve it?
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Post by CtraK on Dec 3, 2011 13:39:07 GMT -5
Said quadratic might be 3x2 + 12, but it's a long, long time since I last did this stuff and I used to suck at it anyway.
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Post by Undecided on Dec 3, 2011 14:06:47 GMT -5
Easiest way to do this is to rewrite the integrand in terms of u= x^2+3 and the measure dx in terms of du=d(x^2+3)=2x dx. The denominator x^4+6x^2+10 is irreducible over the rationals and has a factorization over the reals which involves nested radicals, so attempting a partial fraction decomposition is not in your best interest.
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Post by Sleepy on Dec 3, 2011 16:41:06 GMT -5
I know how to do substitution, but where is your u = x^2 + 3 coming from?
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Post by Old Viking on Dec 3, 2011 16:45:12 GMT -5
A dramatic revelation of why I was an English major.
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Post by Undecided on Dec 3, 2011 18:04:07 GMT -5
- Your integral is an odd rational function f(x), that is, it is a quotient of polynomials and f(-x)= -f(x).
- If a function is an odd rational function, then it is equal to x times an even rational function, say, g(x). Then f(x) = x g(x).
- Now, it is the case that every even rational function g(x) is a rational function in x^2, say h(x): g(x)=h(x^2).
- Thus, the form you are integrating is f(x)dx=x g(x)dx=x h(x^2) dx.
- Making the substitution v = x^2, dv = 2x dx simplifies the problem: f(x)dx=h(v) dv/2.
- It turns out, in your case, that h(v) is quadratic in the denominator, and hence may be solved by completing the square. Let u=v+3. Then k(u)=h(u-3)=h(v) has no linear terms in u.
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Post by Admiral Lithp on Dec 3, 2011 18:19:03 GMT -5
Kill math.
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Post by Sleepy on Dec 3, 2011 18:35:12 GMT -5
- Your integral is an odd rational function f(x), that is, it is a quotient of polynomials and f(-x)= -f(x).
- If a function is an odd rational function, then it is equal to x times an even rational function, say, g(x). Then f(x) = x g(x).
- Now, it is the case that every even rational function g(x) is a rational function in x^2, say h(x): g(x)=h(x^2).
- Thus, the form you are integrating is f(x)dx=x g(x)dx=x h(x^2) dx.
- Making the substitution v = x^2, dv = 2x dx simplifies the problem: f(x)dx=h(v) dv/2.
- It turns out, in your case, that h(v) is quadratic in the denominator, and hence may be solved by completing the square. Let u=v+3. Then k(u)=h(u-3)=h(v) has no linear terms in u.
I follow your first couple bullets, but my mind goes downhill from there. I don't recall ever seeing the points you're raising in my class (unless they're the same thing written differently), though with your v and dv, it looks like integration by parts. (Please don't stab me.)
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Post by Undecided on Dec 3, 2011 18:54:53 GMT -5
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Khris
Full Member
Looks older than they are
Posts: 225
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Post by Khris on Dec 3, 2011 19:18:00 GMT -5
Is that your response to every thing to kill it?
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Post by The_L on Dec 3, 2011 20:21:25 GMT -5
Is that your response to every thing to kill it? Pretty much. Hey Lithp, go to the result of integrating 1/(cabin) d(cabin), - C.
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Post by Sleepy on Dec 4, 2011 11:50:28 GMT -5
Wow, I would've never thought of that. Definitely not something I've seen in class, but I suppose the professor has to accept it. Thanks!
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Post by Yla on Dec 6, 2011 7:50:20 GMT -5
I realize you're done already, but here's how I would go at it: A lot of the times you have some ugly function, you can do a substitution, and the leftover from the transformation (fuck me how it's called) eliminates the ugly bit from the function. If you substitute u = x^2, you have du/dx = 2x <=> dx = du/2x. The x in the enumerator and the new 2x eliminate each other, and 1/2(u^2+6u+10) remains. Something like 1/(x^2 + a) often goes with a trigonometric function, so I would look in a reference for something familiar. Though I don't know whether that would have worked here.
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Post by Sleepy on Dec 6, 2011 11:46:46 GMT -5
Yep, you're certainly right. That's the method Undecided used. I showed it to the professor and he said the answer was right (it ends up being an inverse tangent), but that the problem is completely unfair because of that difficult substitution. So, thankfully similar problems are not fair game for the final exam.
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