A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
find integral of 7x 11 over x^210x + 41
anonymous
 one year ago
find integral of 7x 11 over x^210x + 41

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\\[\int\limits_{}^{} \frac{ 7x11 }{ x^210x+41 }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Is this your integral?\[\Large \int\limits_{}^{}\frac{ 7x11 }{ x^210x+41 }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes. that is correct.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You have to do what is called a usubstitution for this

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Because 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(x5). So what I want to do is make the numerator become 7x35, that way I can hav 7(x5) and get a usubstitution that will work. Kind of get the idea I'm aiming for?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So I can add and subtract 24 to the numerator right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0since it's like adding 0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright, 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: \[\int\limits_{}^{}\frac{ 7x11 }{ x^{2}  10x + 41 }dx = \int\limits_{}^{}\frac{ 7x 1124 + 24 }{ x^{2}  10x+41 }dx = \int\limits_{}^{}\frac{ 7x35 }{ x^{2} 10x + 41 }dx + \int\limits_{}^{}\frac{ 24 }{ x^{2}10x+41 }dx\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry that got cut off

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So with these integrals, do you think you're okay from here, or still unsure what to do next? \[\int\limits_{}^{}\frac{ 7x35 }{ x^{2}10x+41 }dx + \int\limits_{}^{}\frac{ 24 }{ x^{2}10x+41 }dx\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm still kinda confused about where I got from here. Sorry.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, the left integral is a usubstitution. 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(x5)dx dx = du/[2(x5)] we can substitute into our integral and get the cancellations that will let u do the integral. Does that part make sense?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright. So that will give us: \[\int\limits_{}^{}\frac{ 7x35 }{ x^{2}10x+41 }dx = \int\limits_{}^{}\frac{ 7(x5) }{ u }*\frac{ du }{ 2(x5)} = \frac{ 7 }{ 2 }\int\limits_{}^{}\frac{ 1 }{ u }du\] I assume you know the antiderivative from there :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh! Now that makes sense.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright, 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah I've seen it before.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So to get 1 in the numerator i pull out 24.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Right, 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i haven't done that in forever. I remember doing it a while back though.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, 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 \(x^{2} 10x + 25 + 16\) \((x5)^{2} + 16\) Follow that okay? Basically just a refresher in completing the square :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now it all makes sense. Thank you so much, you've saved me.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You're welcome, glad to help :)
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.