Partial Fraction Decomp. Integral

In summary, the conversation is about solving the integral \int\frac{4y^2-7y-12}{y(y+2)(y-3)} with limits of integration [1, 2]. The attempt at a solution involves partial fraction decomposition and finding the constants A, B, and C. The final answer is 2ln(2)+ln(\frac{4}{3})-ln(2), which is not equal to either answer provided in the book. There is a discussion about checking the work and getting the correct values for A, B, and C.
  • #1

Homework Statement



[tex]\int\frac{4y^2-7y-12}{y(y+2)(y-3)}[/tex] with limits of integration [tex][1, 2][/tex]

Homework Equations



The answer given in the book is [tex]\frac{27}{5}ln(2)-\frac{9}{5}ln(3)[/tex]

or [tex]\frac{9}{5}ln\frac{8}{3}[/tex]


The Attempt at a Solution



[tex]\int\frac{4y^2-7y-12}{y(y+2)(y-3)}[/tex]

[tex]=\int\frac{A}{y}+\frac{B}{y+2}+\frac{C}{y-3}[/tex]

I found:

[tex]4y^2-7y-12=A(y+2)(y-3)+B(y)(y-3)+C(y)(y+2)[/tex]

If y=-2, then [tex]B=1[/tex]

I also found:

[tex]4y^2-7y-12=(A+B+C)y^2+(2C-3B-A)y-6A[/tex]

so [tex]6A=12[/tex], or [tex]A=2[/tex].

Also, [tex](A+B+C)=4[/tex] so [tex](2+1+C)=4[/tex] and [tex]C=1[/tex]

Therefore: [tex]=\int\frac{A}{y}+\frac{B}{y+2}+\frac{C}{y-3}[/tex]

[tex]=\int\frac{2}{y}+\frac{1}{y+2}+\frac{1}{y-3}[/tex]

[tex]=2ln|y|+ln|y+2|+ln|y-3|[/tex] with limits [tex][1, 2][/tex]

[tex]=2(ln(2)-ln(1))+(ln(2+2)-ln(1+2)+(ln|2-3|-ln|1-3|)[/tex]

[tex]=2ln(2)+(ln(4)-ln(3)+(ln(1)-ln(2))[/tex]

[tex]=2ln(2)+ln(\frac{4}{3})-ln(2)[/tex]

This is not equal to either answer provided in the book but I can't find an error. Also, how do you put the limits of integration into the function using Latex?
 
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  • #2
Check your work. I get A = 2, B = 9/5, and C = 1/5.
 
  • #3
How do you get B=1? I get B=9/5. Also note you can get A (by putting y=0) and C (by putting y=3) the same way you got B.
 

1. What is partial fraction decomposition (PFD)?

Partial fraction decomposition is a method used in calculus to break down a rational function into simpler fractions. This can make it easier to integrate and solve for the unknown variables.

2. Why is partial fraction decomposition important?

PFD is important because it allows us to solve integrals that would otherwise be difficult or impossible to solve. It also helps us to better understand the behavior of rational functions.

3. How do you perform partial fraction decomposition?

To perform PFD, the rational function must first be written in the form of (a polynomial)/(a polynomial). Then, the numerator is factored and the resulting fractions are set equal to the original function. Finally, the unknown coefficients are solved for using algebraic equations.

4. What types of rational functions can be decomposed using PFD?

PFD can be used on proper rational functions, which are those where the degree of the numerator is less than the degree of the denominator. It can also be used on improper rational functions, but additional steps are required.

5. Are there any limitations to PFD?

Yes, there are limitations to PFD. It can only be used on rational functions, and it may not always result in a simpler form of the integral. In some cases, the decomposition may lead to complex fractions which can be difficult to integrate.

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