How to use fractional decomposition to integrate rational functions?

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Homework Help Overview

The discussion revolves around the integration of a rational function using fractional decomposition, specifically the integral of the form ∫(2x)/(3x² + 10x + 3) dx. Participants are exploring methods to approach this integration problem, particularly focusing on the setup for partial fraction decomposition.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the initial attempts at integration, including considerations of U-substitution and trigonometric substitution. The original poster expresses confusion about fractional decomposition and how to set it up correctly. There are questions about how to equate coefficients after expanding the right-hand side of the equation.

Discussion Status

Some participants have provided guidance on expanding the right side of the equation and collecting terms to solve for the constants A and B. There is an ongoing exploration of how to set up and solve the resulting system of equations, with some participants confirming the correctness of the partial fraction decomposition derived.

Contextual Notes

Participants note that fractional decomposition has not yet been covered in their course, which adds to the complexity of the discussion. There is also a mention of checking the correctness of the decomposition by recombining the fractions.

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Homework Statement


[tex]\int \frac{2x}{3x^{2}+10x+3} dx[/tex]

Homework Equations



The Attempt at a Solution



I can't think of a U-substitution that would work, nor a trigonometric substitution, or integration by part.

[tex]\int \frac{2x}{3x^{2}+10x+3} dx[/tex]
[tex]\int \frac{2x}{(x+3)(3x+1)} dx[/tex]

I factored the denominator out thinking that I could somehow substitute for one product, but that doesn't work clearly. How do you integrate functions like these??

I popped it into wolfram and it had a step about fractional decomposition, but I am having a hard time understanding it and we have not covered it yet in my course.

Here is my go at it:
It has to be in this form right?
[tex]\frac{2x}{(3+x)(3x+1)} = \frac{A}{3+x} + \frac{B}{3x+1}[/tex]

So now I would multiply the LCD through the equation leaving:
[tex]2x = A(3x+1) + B(3+x)[/tex]

I don't understand what to do now though?
 
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Multiply out the right side, and then factor out x from all possible terms.
 
Well, it seems like, in this partial fraction decomposition, you can expand the right hand side of your last equation, collect terms, and then solve for A and B.
 
George Jones said:
Multiply out the right side, and then factor out x from all possible terms.

[tex]2x = A(3x+1) + B(3+x)[/tex]
[tex]2x = x(3A+B) + A +3B[/tex]

?
 
left = right.

How many x's on the left? On the right?

What is the constant on the left? On the right?
 
George Jones said:
left = right.

How many x's on the left? On the right?

What is the constant on the left? On the right?

So,

The constant is A + 3B

The other equation is 2=(3A+B) ?

I'm guessing I system of equation these guys to find A and B now? What does the constant equal? 0?
 
That makes this I believe:

[tex]\frac{2x}{(3+x)(3x+1)} = \frac{\frac{3}{4}}{3+x} + \frac{\frac{-1}{4}}{3x+1}[/tex]

Does that look correct? I can integrate those.
 
Yeah, it seems like you've got it. You can always check your answer by doing the reverse (combine the two terms on the right hand side into one fraction with a common denominator and check that the numerator simplifies to 2x).
 

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