How Do You Solve Integrals Using Partial Fractions?

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SUMMARY

The discussion focuses on solving integrals using partial fractions, specifically the integral $$\int\frac{3x^2+x+12}{(x^2+5)(x-3)}$$. Participants clarify that the correct form for the partial fraction expansion should have numerators of degree one less than the denominators, leading to the expression $$\frac{Ax + B}{x^2 + 5} + \frac{C}{x - 3}$$. The importance of differentiating the final answer to verify correctness is emphasized, along with the potential to use complex numbers for polynomial factorization. The final integral solution includes terms like $$\frac{\arctan\left(\frac{x}{\sqrt{5}}\right)}{\sqrt{5}} + 3\ln\left(\left|x-3\right|\right) + C$$.

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  • Familiarity with polynomial long division and factorization.
  • Knowledge of complex numbers and their application in polynomial expressions.
  • Experience with logarithmic and arctangent functions in calculus.
NEXT STEPS
  • Study the method of partial fraction decomposition in detail.
  • Learn about polynomial long division and its role in integration.
  • Explore the use of complex numbers in calculus, particularly in polynomial factorization.
  • Practice differentiating integrals to verify solutions in calculus.
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Students and educators in calculus, mathematicians working with integrals, and anyone interested in mastering partial fraction decomposition techniques.

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

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