How Do You Solve Integrals Using Partial Fractions?

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Discussion Overview

The discussion centers on solving integrals using partial fractions, specifically focusing on the integral of a rational function involving a quadratic and a linear factor in the denominator. Participants explore the correct form of partial fraction decomposition and the implications of polynomial degrees in this context.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents an integral and proposes a partial fraction decomposition, suggesting coefficients A and B based on specific values of x.
  • Another participant challenges the initial decomposition, stating that the degree of each numerator must be one degree less than the degree of the denominator, proposing a different form with a linear numerator for the quadratic denominator.
  • A further reply clarifies the requirement for linear numerators over quadratic denominators and constants over linear factors, introducing the possibility of using complex numbers for factorization.
  • One participant provides a link to a project related to the discussion, inviting comments and suggestions.

Areas of Agreement / Disagreement

Participants express disagreement regarding the correct form of the partial fraction decomposition, with multiple competing views on the appropriate structure of the numerators and the use of complex numbers.

Contextual Notes

There are limitations regarding the assumptions made about polynomial factorization and the use of complex numbers, which remain unresolved in the discussion.

karush
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partial fractions
$$\int\frac{3x^2+x+12}{(x^2+5)(x-3)}
=\frac{A}{(x^2+5)}+\frac{B}{(x-3)}$$
$$3x^2+x+12=A(x-3)+B(x^2+5)$$
x=3 then 27+3+12=14B
3=B
x=0 then
12=-3A+15
1=A
$$\int\frac{1}{(x^2+5)} \, dx
+3\int\frac{1}{(x-3)}\, dx$$ $\displaystyle
\frac{\arctan\left(\frac{x}{\sqrt{5}}\right)}{\sqrt{5}}
+3\ln\left(\left|x-3\right|\right)+C$

maybe? not sure
 
Last edited:
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karush said:
partial fractions
$$\int\frac{3x^2+x+12}{(x^2+5)(x-3)}
=\frac{A}{(x^2+5)}+\frac{B}{(x-3)}$$
$$3x^2+x+12=A(x-3)+B(x^2+5)$$
x=3 then 27+3+12=14B
3=B
x=0 then
12=-3A+15
1=A
$$\int\frac{1}{(x^2+5)} \, dx
+3\int\frac{1}{(x-3)}\, dx$$ $\displaystyle
\frac{\arctan\left(\frac{x}{\sqrt{5}}\right)}{\sqrt{5}}
+3\ln\left(\left|x-3\right|\right)+C$

maybe? not sure

Try differentiating your answer to see if you get back to where you started. You should quickly see that it is incorrect.

The degree of each numerator needs to be ONE DEGREE LESS than the degree of the denominator. So that means you need to have $\displaystyle \begin{align*} \frac{A\,s + B}{x^2 + 5} + \frac{C}{x - 3} \end{align*}$ as your partial fraction expansion.
 
Typo: Prove It meant \frac{Ax+ B}{x^2+ 5}+ \frac{C}{x- 3}.

Also, he said "one degree less than the denominator". Actually you can always factor a polynomial into linear or quadratic factors (with real coefficients) so you need linear numerators, (Ax+ B), over quadratic denominators and constants (C) over linear factors.

If we were willing to use complex numbers, we can write any polynomial as linear factors and we could write this is \frac{A}{x+ 5i}+ \frac{B}{x- 5i}+ \frac{C}{x- 3}. Of course we could then go back to real numbers by rationalizing the denominators: \frac{A}{x+ 5i}\frac{x- 5i}{x- 5i}+ \frac{B}{x- 5i}\frac{x+ 5i}{x+ 5i}= \frac{Ax+ 5iA+ Bx- 5Bi}{x^2+ 5}= \frac{(A+ B)x+ 5i(A- B)}{x^2+ 5}. Here, it will turn out that both A+ B is real and A- B is imaginary so that both A+ B and 5i(A- B) are real.
 
https://dl.orangedox.com/geAQogxCM0ZYQLaUju

a project of a collection of mhb replies (WIP)

Mahalo for comments and suggestions

SSCwt.png
 

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