Question

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

Answer 1

\\[\int\limits_{}^{} \frac{ 7x-11 }{ x^2-10x+41 }\]

Answer 2

Is this your integral?\[\Large \int\limits_{}^{}\frac{ 7x-11 }{ x^2-10x+41 }\]

Answer 3

yes. that is correct.

Answer 4

You have to do what is called a u-substitution for this

Answer 5

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?

Answer 6

I think I follow.

Answer 7

So I can add and subtract 24 to the numerator right?

Answer 8

since it's like adding 0

Answer 9

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:

\[\int\limits_{}^{}\frac{ 7x-11 }{ x^{2} - 10x + 41 }dx = \int\limits_{}^{}\frac{ 7x -11-24 + 24 }{ x^{2} - 10x+41 }dx = \int\limits_{}^{}\frac{ 7x-35 }{ x^{2} -10x + 41 }dx + \int\limits_{}^{}\frac{ 24 }{ x^{2}-10x+41 }dx\]

Answer 10

Sorry that got cut off

Answer 11

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.

Answer 12

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

Answer 13

I'm still kinda confused about where I got from here. Sorry.

Answer 14

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?

Answer 15

Yes.

Answer 16

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 :)

Answer 17

Oh! Now that makes sense.

Answer 18

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.

Answer 19

Yeah I've seen it before.

Answer 20

So to get 1 in the numerator i pull out 24.

Answer 21

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?

Answer 22

i haven't done that in forever. I remember doing it a while back though.

Answer 23

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 :)

Answer 24

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?

Answer 25

Bingo :)

Answer 26

Now it all makes sense. Thank you so much, you've saved me.

Answer 27

You're welcome, glad to help :)