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anonymous

  • one year ago

find integral of 7x -11 over x^2-10x + 41

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  1. anonymous
    • one year ago
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    \\[\int\limits_{}^{} \frac{ 7x-11 }{ x^2-10x+41 }\]

  2. anonymous
    • one year ago
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    Is this your integral?\[\Large \int\limits_{}^{}\frac{ 7x-11 }{ x^2-10x+41 }\]

  3. anonymous
    • one year ago
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    yes. that is correct.

  4. anonymous
    • one year ago
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    You have to do what is called a u-substitution for this

  5. anonymous
    • one year ago
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    Because the numerator is one degree less than the denominator, this trick will work. We're going to do a manipulation in order to make the derivative of the denominator appear in the numerator. So if we were to choose u to be x^2 - 10x + 41, we would get 2x - 10 = 2(x-5). So what I want to do is make the numerator become 7x-35, that way I can hav 7(x-5) and get a u-substitution that will work. Kind of get the idea I'm aiming for?

  6. anonymous
    • one year ago
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    I think I follow.

  7. anonymous
    • one year ago
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    So I can add and subtract 24 to the numerator right?

  8. anonymous
    • one year ago
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    since it's like adding 0

  9. anonymous
    • one year ago
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    Alright, cool. So we currently have 7x - 11 and I want 7x - 35. So what I'm going to do is add and subtract the same quantity, 24, and then split this into two integrals: \[\int\limits_{}^{}\frac{ 7x-11 }{ x^{2} - 10x + 41 }dx = \int\limits_{}^{}\frac{ 7x -11-24 + 24 }{ x^{2} - 10x+41 }dx = \int\limits_{}^{}\frac{ 7x-35 }{ x^{2} -10x + 41 }dx + \int\limits_{}^{}\frac{ 24 }{ x^{2}-10x+41 }dx\]

  10. anonymous
    • one year ago
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    Sorry that got cut off

  11. anonymous
    • one year ago
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    But yes, you have the right idea. That 2nd integral that cut off just has 24 in the numerator. Either way, I now have two integrals I can solve.

  12. anonymous
    • one year ago
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    So with these integrals, do you think you're okay from here, or still unsure what to do next? \[\int\limits_{}^{}\frac{ 7x-35 }{ x^{2}-10x+41 }dx + \int\limits_{}^{}\frac{ 24 }{ x^{2}-10x+41 }dx\]

  13. anonymous
    • one year ago
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    I'm still kinda confused about where I got from here. Sorry.

  14. anonymous
    • one year ago
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    Well, the left integral is a u-substitution. That was the plan all along and the reason we did the manipulation we did. If we do this: u = x^2 -10x + 41 du = 2(x-5)dx dx = du/[2(x-5)] we can substitute into our integral and get the cancellations that will let u do the integral. Does that part make sense?

  15. anonymous
    • one year ago
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    Yes.

  16. anonymous
    • one year ago
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    Alright. So that will give us: \[\int\limits_{}^{}\frac{ 7x-35 }{ x^{2}-10x+41 }dx = \int\limits_{}^{}\frac{ 7(x-5) }{ u }*\frac{ du }{ 2(x-5)} = \frac{ 7 }{ 2 }\int\limits_{}^{}\frac{ 1 }{ u }du\] I assume you know the antiderivative from there :)

  17. anonymous
    • one year ago
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    Oh! Now that makes sense.

  18. anonymous
    • one year ago
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    Alright, awesome. For the 2nd integral, we need this prerequisite knowledge: \[\int\limits_{}^{}\frac{ 1 }{ u^{2} + a^{2} }du = \frac{ 1 }{ a }\arctan(\frac{ u }{ a }) + C\] You may have seen this before.

  19. anonymous
    • one year ago
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    Yeah I've seen it before.

  20. anonymous
    • one year ago
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    So to get 1 in the numerator i pull out 24.

  21. anonymous
    • one year ago
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    Right, you can pull out the 24 if it looks better to ya. In the end, our result will be multiplied by 24 anyway :) So we need to get the integral into a form like that, which we'll do by completing the square. Are you comfortable with completing the square?

  22. anonymous
    • one year ago
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    i haven't done that in forever. I remember doing it a while back though.

  23. anonymous
    • one year ago
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    Yeah, its what we often need to do to get the denominator into a form that will let us do those inverse trig solutions. Well, the main idea is to first factor out the coefficient of the x^2 term from the two variable terms. Which is 1, so no worries there. From there, we take the coefficient of the x to the first power term, half it then square it. This number we create is both added and subtracted in the denominator. So if we do that we get this result: \(x^{2}-10x + 41\) = \(x^{2} - 10x + 25 - 25 + 41\) = \(x^{2} - 10x +25 + 16\) This result always creates a perfect square trinomial that can be factored. The factored expression is x + half the coefficent of the x to the first power term, all squared. If we phrase that in terms of the general quadratic ax^2 + bx + c, then the perfect square trinomial factors into \(a(x+b/2)^{2}\). GIven that, we can instantly factor and get this result \(x^{2} -10x + 25 + 16\) \((x-5)^{2} + 16\) Follow that okay? Basically just a refresher in completing the square :)

  24. anonymous
    • one year ago
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    Oh and then I can just sub x -5 for u and then it's in the form 1 over x^2 + A^2 form where A is 4 right?

  25. anonymous
    • one year ago
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    Bingo :)

  26. anonymous
    • one year ago
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    Now it all makes sense. Thank you so much, you've saved me.

  27. anonymous
    • one year ago
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    You're welcome, glad to help :)

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