How to use fractional decomposition to integrate rational functions?

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SUMMARY

The discussion focuses on the integration of the rational function \(\int \frac{2x}{3x^{2}+10x+3} dx\) using fractional decomposition. The user initially struggles with finding an appropriate substitution method but eventually identifies the need for partial fraction decomposition. The correct form is established as \(\frac{2x}{(3+x)(3x+1)} = \frac{A}{3+x} + \frac{B}{3x+1}\), leading to the equations \(2 = (3A + B)\) and \(A + 3B = 0\). The user successfully determines the values of A and B as \(\frac{3}{4}\) and \(-\frac{1}{4}\), respectively, allowing for the integration of the decomposed fractions.

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  • Understanding of rational functions and their properties.
  • Familiarity with integration techniques, particularly partial fraction decomposition.
  • Basic algebra skills for solving systems of equations.
  • Knowledge of polynomial factorization.
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  • Study the method of partial fraction decomposition in detail.
  • Practice integrating various rational functions using fractional decomposition.
  • Learn how to verify integration results by combining fractions.
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Students studying calculus, particularly those learning about integration techniques, as well as educators seeking to enhance their teaching methods in rational function integration.

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Homework Statement


\int \frac{2x}{3x^{2}+10x+3} dx

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.

\int \frac{2x}{3x^{2}+10x+3} dx
\int \frac{2x}{(x+3)(3x+1)} dx

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?
\frac{2x}{(3+x)(3x+1)} = \frac{A}{3+x} + \frac{B}{3x+1}

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

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.

2x = A(3x+1) + B(3+x)
2x = x(3A+B) + A +3B

?
 
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:

\frac{2x}{(3+x)(3x+1)} = \frac{\frac{3}{4}}{3+x} + \frac{\frac{-1}{4}}{3x+1}

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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