MHB How Do You Solve Integrals Using Partial Fractions?

Click For Summary
The discussion focuses on solving integrals using partial fractions, specifically for the integral of a rational function. The correct approach involves expressing the integrand as a sum of fractions with numerators that are one degree less than their respective denominators. Participants emphasize the need for linear numerators over quadratic denominators and constants over linear factors. There is also a mention of using complex numbers to factor polynomials, which can simplify the process. The conversation highlights the importance of verifying solutions by differentiation to ensure accuracy.
karush
Gold Member
MHB
Messages
3,240
Reaction score
5
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:
Physics news on Phys.org
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
 

Similar threads

Replies
6
Views
2K
Replies
5
Views
2K
Replies
5
Views
2K
Replies
8
Views
2K
Replies
6
Views
2K
Replies
6
Views
3K
Replies
3
Views
2K