Solve Partial Fractions: Step-by-Step Guide

In summary, the conversation discusses how to evaluate the integral of 1/(x^2-1) using partial fractions. The participants suggest different methods, such as solving for A and B, using natural log rules, and substituting values for x. They eventually agree on a simpler method of finding A and B by multiplying the equation by x^2-1 and substituting values for x.
  • #1
LadiesMan
96
0
[SOLVED] Partial Fractions

1. Evaluate:

[tex]\int \frac{dx}{x^{2} -1}[/tex]



Attempt:

[tex]\int \frac{dx}{x^{2} -1}[/tex]

[tex]= \int \frac{dx}{(x+1)(x-1)}[/tex]

[tex]= \frac{A}{(x+1)} + \frac{B}{(x-1)}[/tex]

[tex]= \frac{Ax - A + Bx + B}{(x+1)(x-1)}[/tex]

Where do I got from here? Thanks
 
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  • #2
Hi LadiesMan! :smile:

(erm … what happened to the integral sign, and dx … ?)

Easy! You just solve Ax - A + Bx + B = 1.

So A = … , B = … ? :smile:
 
  • #3
so A = Ax + Bx + B -1
and B = -Ax + A - Bx -1??
 
  • #4
LadiesMan said:
so A = Ax + Bx + B -1
and B = -Ax + A - Bx -1??

Ax-A+Bx+B=1 for all values of x

meaning for any value of x you put in there, it will always be equal to 1
Choose some suitable values for x to get A and B
 
  • #5
You could do A(x-1) + B(x+1) = 1

Then to get rid of the A, substitute x = 1 and find B.

To get rid of the B, substitute x = -1 and find A.
 
  • #6
That means:

(1/2)ln(x-1) - (1/2)ln(x+1) + C

Therefore Since: (1/2)ln(x-1) = ln(x-1)^1/2 and (1/2)ln(x +1)= (1/2)ln(x +1)^1/2

ln(x-1)^1/2 - (1/2)ln(x +1)^1/2 + C

Natural log rule ln a - ln b = ln (a/b)

Thank you =)
 
  • #7
… never substitute …

rock.freak667 said:
Choose some suitable values for x to get A and B

No! :frown:

It's a polynomial, and you just put all the coefficients equal to 0.

It's (A+B)x + (B-A-1) = 0,

so both the brackets must be 0. :smile:

Never substitute - it makes it look as if you don't understand what a polynomial is!

(I agree that, technically, that's the same as substituting x = 0 and ∞; but that requires an extra line, and it doesn't work if you have x^2 or higher.)
 
  • #8
Tiny Tim, since that equation is true for all x, you certainly could get substitute any two values of x to get 2 equations to solve for a and b. Of course, it is true that if a polynomial is 0 for all x, then its coefficients must all be 0- that's typically simpler than substituting specific values for x.

Ladies Man, a somewhat simpler method for partial fraction is this:
You want to find A, B such that
[tex]\frac{A}{x+1}+ \frac{B}{x-1}= \frac{1}{x^2- 1}[/tex]
Multiply both sides by x2-1 to eliminate the denominators:
[tex]A(x-1)+ B(x+ 1)= 1[/itex]
Now substitute x= 1 and x= -1 to find A and B.
 
  • #9
Thanks everyone! =P
 

FAQ: Solve Partial Fractions: Step-by-Step Guide

What is the concept of partial fractions?

Partial fractions is a method of breaking down a complex fraction into simpler fractions. It is used to simplify integration by breaking down a rational function into smaller, easier-to-integrate parts.

What are the steps involved in solving partial fractions?

The steps involved in solving partial fractions are:

  1. Factor the denominator of the rational function into linear and irreducible quadratic factors.
  2. Write the original fraction as a sum of simpler fractions with the same denominator.
  3. Set up a system of equations by equating the numerators of the original fraction and the simpler fractions.
  4. Solve the system of equations to find the unknown coefficients.
  5. Combine the simpler fractions using the found coefficients to obtain the partial fraction decomposition of the original fraction.

What are the different types of partial fractions?

The different types of partial fractions are:

  • Proper fractions - where the degree of the numerator is less than the degree of the denominator.
  • Improper fractions - where the degree of the numerator is greater than or equal to the degree of the denominator.
  • Partial fractions with distinct linear factors - where the denominator can be factored into distinct linear factors.
  • Partial fractions with repeated linear factors - where the denominator can be factored into repeated linear factors.
  • Partial fractions with irreducible quadratic factors - where the denominator cannot be factored further and contains irreducible quadratic factors.

Can partial fractions be used in both indefinite and definite integrals?

Yes, partial fractions can be used in both indefinite and definite integrals. In indefinite integrals, partial fractions help to simplify the integration process. In definite integrals, partial fractions help to evaluate the integral over a specific interval.

Are there any special cases or exceptions when solving partial fractions?

Yes, there are a few special cases or exceptions when solving partial fractions. These include:

  • When the degree of the numerator is equal to or greater than the degree of the denominator, long division must be used before applying partial fractions.
  • When the denominator has repeated linear factors, the coefficients for each factor are different and must be determined separately.
  • When the denominator has a factor with a multiplicity greater than 1, the corresponding coefficient in the partial fraction decomposition will have a higher power in the numerator.
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