Integration by Partial Fractions

In summary, the given integral can be solved by factoring out the denominator and breaking up the main fraction into separate fractions. The coefficients A and B can be found by choosing x values that make the other coefficient go to zero. After breaking up the main integral into a sum of integrals, the fundamental theorem of calculus can be used to find the value. The correct answer is 6ln(3).
  • #1

Homework Statement



[tex]\int_0^1 \! \frac{x-4}{x^2+3x+2} dx[/tex]


Homework Equations





The Attempt at a Solution



Factoring out the denominator...

[tex]x^2+3x+2 = (x+1)(x+2)[/tex]​


Breaking up the main fraction into the separate fractions. Since both are linear, only a single constant is needed on top.

[tex]\frac{A}{x+1}+\frac{B}{x+2}[/tex]​

Fining the coefficients (A and B) by choosing x values that make the other coefficient go to zero.

[tex]x-4 = A(x+2) + B(x+1)[/tex]

[tex]x=-2[/tex]
[tex]-6=B(-1)[/tex]
[tex]6=B[/tex]

[tex]x=-1[/tex]
[tex]-5=A(1)[/tex]
[tex]-5=A[/tex]​


Back in the integral they go. Also, breaking up the main integral into a sum of integrals.

[tex]\int \frac{-5}{x+1}dx + \int \frac{6}{x+2} dx[/tex]

[tex]-5ln(x+1) + 6ln(x+2) + c[/tex]


Now using the fundamental theorem of calculus to find the value.

[tex](-5ln(1+1) + 6ln(1+2))-(-5ln(0+1) + 6ln(0+2))[/tex]​


Result:

[tex]6ln(3)-11ln(2)[/tex]​

Wrong. Aw.
 
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  • #2
It looks OK to me. Who told you the answer was wrong?
 
  • #3
P.S. Wolfram Alpha's answer agrees numerically with yours.
 
  • #4
The online homework system... though it gave me multiple options, and none of them were this. I suppose I could try to figure out which is equal, if my work is indeed correct.

http://mikederoche.com/temp/options.png [Broken]
 
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  • #5
Got it... equivalent to the second option.
 
  • #6
It looks like they really want you to have a [itex]\ln(3/2)[/itex] term for some reason. OK then:

[tex]6 \ln(3) = 6 \ln(2(3/2)) = 6\ln(2) + 6\ln(3/2)[/tex]
 

1. What is Integration by Partial Fractions?

Integration by Partial Fractions is a technique used in calculus to simplify and solve integrals involving rational functions. It involves breaking down a complex fraction into simpler, more manageable fractions.

2. When is Integration by Partial Fractions used?

Integration by Partial Fractions is usually used when the integral involves a rational function, where the numerator and denominator are both polynomials. It is also used when the degree of the numerator is less than the degree of the denominator.

3. How is Integration by Partial Fractions done?

The general steps for Integration by Partial Fractions are as follows:
1. Factor the denominator of the rational function into linear or irreducible quadratic factors.
2. Write the partial fraction decomposition, with each factor of the denominator having its own constant coefficient.
3. Set up a system of equations using the original rational function and the partial fraction decomposition.
4. Solve the system of equations to find the coefficients.
5. Substitute the coefficients into the partial fraction decomposition.
6. Integrate each term of the partial fraction decomposition.
7. Combine the integrated terms to obtain the final solution.

4. Why is Integration by Partial Fractions useful?

Integration by Partial Fractions allows us to solve integrals that would otherwise be difficult or impossible to solve using other methods. It also helps us to simplify complex integrals, making them easier to evaluate.

5. Are there any limitations to Integration by Partial Fractions?

Integration by Partial Fractions can only be used when the denominator of the rational function can be factored into linear or irreducible quadratic factors. It also may not work for all rational functions, and sometimes the system of equations needed to solve for the coefficients can be difficult to set up and solve.

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