integral of (3x+5)/(5x^2-4x+1)dx

- anonymous

integral of (3x+5)/(5x^2-4x+1)dx

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- schrodinger

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- anonymous

It's a rational polynomial, have you tried factoring it?

- anonymous

let me do it real quick

- anonymous

it cant be factored..

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- anonymous

imaginary roots?

- anonymous

huh?

- anonymous

imaginary roots wont help you in this i think... hold on... (props for a challenging question)

- anonymous

You said it can't be factored... what are the roots of the denominator?

- anonymous

2/5-i/5, 2/5+i/5?

- anonymous

rational functions like this often don't have elementary anti derivatives.

- anonymous

Is this a homework problem?

- anonymous

it's a practice problem for my exam this coming thursday. I'm trying to study and get familiar with the problems.

- anonymous

ok got it, in this type of question you need to rewrite it so that at least one of the parts of the numerator is equal to the derivative of the denominator.
So we may rewrite it like so: 31/5 integral(1/5x^2-4x+1) +3/10 integral((10x-4)/(5x^2-4x+1))
Use u substitution for integral((10x-4)/(5x^2-4x+1))
u= the denominator and du = the numerator
So now you get 3/10 1/u du (which is also the same as 3/10 ln(u)
Now substitute x back into u
Now, rewrite the other part of the integral (complete the square) to find an inverse tan function hidden there. It simplifies to the equation 31/5 (arctan(2-5x) so the final equation is :
3/10 ln(5x^2-4x+1) + 31/5 (arctan(2-5x) +c (do NOT forget the +C after all this work lol)

- anonymous

if you want a more insightful thought process of a certain part of a problem, please ask (i was kinda just in a hurry to get it done)

- anonymous

woah woah, explain how you rewrote it.. i dont get that..

- anonymous

Okay, so, we know the derivative of the denominator is 10x-4 right? so we just have to do a lot of algebra to get at least one part of the numerator to look like that (its mostly a lucky guess in my case) because i had in the back of my mind that a problem as complex as this needs to use u substitiution

- anonymous

so, i have to get the numerator to be 10x-4? (which is the derivative of the denom)

- anonymous

3x+5 = 31/5 + 3/10 (10x-4) ; (the 2 numerators of the integrals before i factored out constants

- anonymous

Yes, in a problem where there is a denominator, you need to first think of substitution and how you can get the derivative of the denominator into the numerator of the function

- anonymous

I still don't get how in the worl you get 31/5+3/10(10x-4)... out of 3x+5

- anonymous

ok, so, we need to get 10x-4 out of this so : 3x+5 = 30x+50, then subtract 3(10x-4) to get a remainder of 62 (this will go in the 2nd integral)

- anonymous

keep in mind, this is all over a denominator of 10 (which we inserted into the equation when we multiplied 3x+5 by 10)

- anonymous

so multiply by ten and subtract 3(10x-5)?

- anonymous

yeah, and hope you get lucky later on ; )

- anonymous

how do you subtract 3(10x-5) from 30x+50? lol! im so sorry... i just really dont understand

- anonymous

no problem, ive been told i make a horrible teacher XD (and it's 3(10x-4)
So, 30x+50 - 3(10x-4) = 30x+50 - 30x + 12 which makes 62 (the constant in the 2nd integral) then you simply divide by 10 to get 31/5 which is what i factored out of the integral

- anonymous

ahhh, got it,. what you mean constant in the 2nd integral,. ?

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