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

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

Mathematics
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\\[\int\limits_{}^{} \frac{ 7x-11 }{ x^2-10x+41 }\]
Is this your integral?\[\Large \int\limits_{}^{}\frac{ 7x-11 }{ x^2-10x+41 }\]
yes. that is correct.

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You have to do what is called a u-substitution for this
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?
I think I follow.
So I can add and subtract 24 to the numerator right?
since it's like adding 0
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: \[\int\limits_{}^{}\frac{ 7x-11 }{ x^{2} - 10x + 41 }dx = \int\limits_{}^{}\frac{ 7x -11-24 + 24 }{ x^{2} - 10x+41 }dx = \int\limits_{}^{}\frac{ 7x-35 }{ x^{2} -10x + 41 }dx + \int\limits_{}^{}\frac{ 24 }{ x^{2}-10x+41 }dx\]
Sorry that got cut off
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.
So with these integrals, do you think you're okay from here, or still unsure what to do next? \[\int\limits_{}^{}\frac{ 7x-35 }{ x^{2}-10x+41 }dx + \int\limits_{}^{}\frac{ 24 }{ x^{2}-10x+41 }dx\]
I'm still kinda confused about where I got from here. Sorry.
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?
Yes.
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 :)
Oh! Now that makes sense.
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.
Yeah I've seen it before.
So to get 1 in the numerator i pull out 24.
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?
i haven't done that in forever. I remember doing it a while back though.
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 :)
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?
Bingo :)
Now it all makes sense. Thank you so much, you've saved me.
You're welcome, glad to help :)

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