Tricky integral using Partial Fractions

In summary, to evaluate the integral of (13x-4)/(6x^2-x-2) dx, the integrand can be expressed in partial fractions. It is helpful to remove the linear term from the numerator before starting, and the integral will be equal to 13/12 Log[6x^2-x-2] plus the integral of -35/12 1/(6x^2-x-2).
  • #1
thomas49th
655
0

Homework Statement


The question asks me to express the integrand in partial fractions to evaluate the integral

[tex] \int \frac{13x-4}{6x^{2} -x -2} dx [/tex]

Homework Equations





The Attempt at a Solution



Well 6x² -x - 2 doesn't factorise (or I can't see it factorised).

So I tried doing long polynomial division and I got something that looked nothing like what the answer is. How should I begin to tackle this problem?

Thanks
Thomas
 
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  • #3
6x² -x - 2 = 1/6 (6x + 3)(6x - 4)
 
  • #4
And as Gregg suggested you can remove the term linear in x (i.e. 13 x) term from the numerator before you start. That will save you work on the partial fraction expansion:

You can write: 13 = 12 * (13/12)

The numerator can thus be written as:

13/12 (12 x - 48/13) =

13/12 (12 x - 1 + 1 - 48/13) =

13/12 (derivative of denominator -35/13) =

13/12 derivative of denominator - 35/12

So, the integral will be 13/12 Log[6 x^2 - x -2] plus the integral of

-35/12 1/(6 x^2 - x -2)
 

FAQ: Tricky integral using Partial Fractions

What is the concept behind using Partial Fractions to solve tricky integrals?

The concept behind using Partial Fractions is to break down a complex fraction into simpler fractions that can be integrated easily. This is done by expressing the complex fraction as a sum of simpler fractions with unknown constants in the numerator.

How do you determine the unknown constants in Partial Fractions?

The unknown constants in Partial Fractions can be determined by finding a common denominator and equating the coefficients of the terms on both sides of the equation. This creates a system of equations that can be solved to find the values of the unknown constants.

Can Partial Fractions be used to solve all types of integrals?

No, Partial Fractions can only be used to solve integrals where the denominator can be factored into linear or quadratic terms. If the denominator has higher degree terms or irreducible factors, other techniques such as integration by parts or substitution may be required.

Are there any special cases to consider when using Partial Fractions?

Yes, there are a few special cases to consider when using Partial Fractions. These include repeated linear factors, irreducible quadratic factors, and complex roots. In these cases, additional steps may be required to properly decompose the fraction and solve the integral.

Can Partial Fractions be applied to improper integrals?

Yes, Partial Fractions can be used to solve improper integrals as long as the denominator can be factored into linear or quadratic terms. In cases where the improper integral does not converge, Partial Fractions may not be applicable.

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