Integration by partial fractions

In summary: I really appreciate your contribution, but this conversation is about integration by partial fractions and not simplification of algebraic expressions. In summary, the conversation includes a question about solving a specific type of integration problem using partial fractions and different methods are discussed, including the standard method and the heathside method. The conversation also includes a step-by-step solution to the problem using the heathside method.
  • #1
clope023
992
131
[SOLVED] integration by partial fractions

Homework Statement




[tex]\int[/tex]((2x^2-1)/(4x-1)(x^2+1))dx

Homework Equations



A1/ax+b + A2/(ax+b)^2 + ... + An/(ax+b)^n

The Attempt at a Solution



(2x^2-1)/(4x-1)(x^2+1) = A/4x-1 + Bx+C/x^2+1

2x^2-1/x^2+1 = A + Bx+C(4x-1)/x^2+1

set x = 1/4, get

-14/17 = A + B(0), A = -14/17

I'm pretty much stuck here; I just need to get the values for B and C and I can solve it from there, any help is greatly appreciated.
 
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  • #2
I've found the value for C:

2x^2-1/x^2+1 = -14/17 + 4Bx^2-Bx+4Cx-C/x^2+1

set x = 0, get

-1 = -14/17 -C, -3/17 = -C, C = 3/17
 
  • #3
I find it more usefull to put the right hand side over a common denominator
[tex]{A(x^{2}+1) + (Bx+C)(4x-1)}\over{(4x-1)(x^{2}+1)}[/tex].

Now expand the numerator on both sides of the equation (i.e. your first equation in part 3.) and group the like powers of x.
Then the coefficient of each power must match on either side of the equal sign. Matching the coefficients should give you a system of 3 equation that you can solve for A, B, and C.
 
  • #4
Pacopag said:
I find it more usefull to put the right hand side over a common denominator
[tex]{A(x^{2}+1) + (Bx+C)(4x-1)}\over{(4x-1)(x^{2}+1)}[/tex].

Now expand the numerator on both sides of the equation (i.e. your first equation in part 3.) and group the like powers of x.
Then the coefficient of each power must match on either side of the equal sign. Matching the coefficients should give you a system of 3 equation that you can solve for A, B, and C.

yeah I know there are 2 methods to do integration by partial fractions, the standard method which you're describing and another method called heathside which I've been using. I'm more comfortable with the standard method but the heathside method seems to be working better for the problems my professor assigned.

I tried the grouping like powers of x, but it left me with this:

2x^2-1 = Ax^2 + A + 4Bx^2-Bx+4Cx - C

it left me with the -Bx + 4Cx with no like power to go along with.
 
  • #5
When the corresponding power is missing, that means the coefficient is zero.
[tex]2x^{2}-1=2x^{2}+0x-1[/tex].
The coefficient of x is zero, so 4C-B=0 would be your missing equation.
 
  • #6
wow! I never realized that; the damn book doesn't explain specific cases like that; thanks for the help, I'll post the result here in a minute.
 
  • #7
wow, I hate this section:

[tex]\int[/tex]((-14/17)/(4x-1))dx + [tex]\int[/tex]((12/17x + 3/17)/(x^2+1))dx

for the 1st integral:

u = 4x - 1

du = 4dx or 1/4du = dx

so (-14/17)/4[tex]\int[/tex](du/u) = -7/34ln(4x-1)

for the 1st half of the 2nd integral

[tex]\int[/tex](12/17x/x^2+1)dx

u = x^2+1, du = 2xdx or 1/2du = xdx

--> (12/17)/2[tex]\int[/tex]du/u = 6/17ln(x^2+1)

for the last half of the 2nd integral

3/17[tex]\int[/tex](dx/x^2+1) = arctanx + C

so the whole thing equals out to:

-7/34ln(4x-1) + 6/17ln(x^2+1) + 3/17arctanx + C

thanks for the help man.
 
  • #8
Personally, I'd simplify it first by saying:
[tex](2x^2\,-\,1)/(4x\,-\,1)(x^2\,+\,1)[/tex]
[tex]=\,((2x^2\,+\,2)\,-\,3)/(4x\,-\,1)(x^2\,+\,1)[/tex]
[tex]=\,2/(4x\,-\,1)\,-\,3/(4x\,-\,1)(x^2\,+\,1)[/tex]​
 

What is integration by partial fractions?

Integration by partial fractions is a method used in calculus to integrate rational functions that cannot be easily integrated using other methods. It involves breaking down a complex fraction into simpler fractions that can be integrated separately.

Why is integration by partial fractions useful?

Integration by partial fractions allows us to solve integrals that would otherwise be impossible or very difficult to solve. It is particularly useful when dealing with rational functions that involve polynomials of higher degree.

What are the steps involved in integration by partial fractions?

The first step is to factor the denominator of the given rational function into linear and irreducible quadratic factors. Then, set up the partial fraction decomposition by writing the numerator in terms of unknown constants. Next, use algebraic manipulation to find the values of the constants. Finally, integrate each term separately to obtain the final answer.

Can any rational function be integrated using partial fractions?

No, not all rational functions can be integrated using partial fractions. The method only works for proper rational functions, which are functions where the degree of the numerator is less than the degree of the denominator. If the rational function is improper, it must first be simplified using polynomial long division before integration by partial fractions can be applied.

Are there any limitations or special cases in integration by partial fractions?

Yes, there are a few limitations and special cases to consider when using integration by partial fractions. These include repeated linear factors, repeated irreducible quadratic factors, and improper rational functions. Each of these cases requires a slightly different approach in the method, and it is important to carefully consider these before attempting to integrate using partial fractions.

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