Integration of Rational Functions

In summary, the conversation is about evaluating the integral ∫ds/(s^2(s − 1)^2) using the method of partial fractions. The individual attempted a solution but realized there was an error in their partial fraction decomposition. After recalculating the parameters, they were able to find the correct values and solve the integral.
  • #1
forestmine
203
0

Homework Statement



Evaluate the integral. (Remember to use ln |u| where appropriate.)

∫ds/s^2(s − 1)^2


Homework Equations





The Attempt at a Solution



I attempted a solution using the method of partial fractions, but it seems my answer is wrong. Here's what I did...

1=A/s +B/s^2+C/(s-1)+D/(s-1)^2

Then multiplying by a common denominator,

1=As(s-1)^2 +B(s-1)^2 +Cs^2(s-1) + Ds^2
1=A(s^3-2s^2+s) + B(s^2 - 2s +1) + C(s^3 -s^2) +Ds^2

Equating the coefficients and solving for A, B, C, and D, I get A=1, B=1, C=-1, D=3

So my integral now looks like

∫1/s+1/s^2-1/(s-1)+3/(s-1)^2

And taking the integral, I got

lns - 1/s - ln(s-1) + 3/(s-1)

which evidently is wrong.

If anyone can point me in the right direction in terms of where I went wrong, it would be greatly appreciated.

Thanks!
 
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  • #2
forestmine said:

Homework Statement



Evaluate the integral. (Remember to use ln |u| where appropriate.)

∫ds/(s^2(s − 1)^2)

Homework Equations



The Attempt at a Solution



I attempted a solution using the method of partial fractions, but it seems my answer is wrong. Here's what I did...

1=A/s +B/s^2+C/(s-1)+D/(s-1)^2

Then multiplying by a common denominator,

1=As(s-1)^2 +B(s-1)^2 +Cs^2(s-1) + Ds^2
1=A(s^3-2s^2+s) + B(s^2 - 2s +1) + C(s^3 -s^2) +Ds^2

Equating the coefficients and solving for A, B, C, and D, I get A=1, B=1, C=-1, D=3

So my integral now looks like

∫1/s+1/s^2-1/(s-1)+3/(s-1)^2

And taking the integral, I got

lns - 1/s - ln(s-1) + 3/(s-1)

which evidently is wrong.

If anyone can point me in the right direction in terms of where I went wrong, it would be greatly appreciated.

Thanks!
There is an error in your partial fraction decomposition. D is incorrect.
 
  • #3
Thanks for the reply!

So for s^2, I got -2A - 2B - C + D = 0

plugging in A, B, and C:

-2(1) - 2(1) - (-1) = 0

and I get D=3?

I'm not seeing where I'm going wrong...
 
  • #4
Recalculate your parameters and show your work. (I got A=2, B=1, C=-2, D=1. )

ehild
 
  • #5
Wow, I wonder if I'm going about this all wrong.

Here's what I did:

1=As(s-1)^2 +B(s-1)^2 +Cs^2(s-1) + Ds^2
1=A(s^3-2s^2+s) + B(s^2 - 2s +1) + C(s^3 -s^2) +Ds^2

For x^0 = 1=B

For x = 0 = A + B ----> A=-B=-1 (But that contradicts your answer)

For x^2 = 0 = -2A -2B - C + D

For x^3 = 0 = A + C ----> C=1

Looks like all of my values are wrong. What am I doing incorrectly?
 
  • #6
forestmine said:
Wow, I wonder if I'm going about this all wrong.

Here's what I did:

1=As(s-1)^2 +B(s-1)^2 +Cs^2(s-1) + Ds^2
Where does the above line come from.

Seems like a strange place to start --- like you're skipping some steps.
 
  • #7
It comes from here:

1/s^2(s − 1)^2=A/s +B/s^2+C/(s-1)+D/(s-1)^2

And then multiplying all of the above by s^2(s − 1)^2
 
  • #8
forestmine said:
Wow, I wonder if I'm going about this all wrong.

Here's what I did:

1=As(s-1)^2 +B(s-1)^2 +Cs^2(s-1) + Ds^2
1=A(s^3-2s^2+s) + B(s^2 - 2s +1) + C(s^3 -s^2) +Ds^2

For x^0 = 1=B

For x = 0 = A + B ----> A=-B=-1 (But that contradicts your answer)

For s^1: A-2B=0 A=-2B=-2.


ehild
 
  • #9
Got it -- I was in fact equating the coefficients incorrectly.

Thanks so much for all the help!
 

Related to Integration of Rational Functions

What is integration of rational functions?

Integration of rational functions is a mathematical process of finding the antiderivative or indefinite integral of a rational function. It involves reversing the process of differentiation and finding a function whose derivative is the given rational function.

What is a rational function?

A rational function is a function in which the numerator and denominator are both polynomials. It is written in the form f(x) = P(x)/Q(x), where P(x) and Q(x) are polynomials and Q(x) is not equal to 0.

What is the general procedure for integrating rational functions?

The general procedure for integrating rational functions involves breaking down the rational function into simpler fractions, finding the antiderivative of each fraction, and then adding the resulting antiderivatives together. This is known as the method of partial fractions.

What is the difference between definite and indefinite integrals?

A definite integral has limits of integration, which specify the range of values for the independent variable. It gives a numerical value as the result. On the other hand, an indefinite integral has no limits of integration and gives a general expression as the result.

What are some common techniques for integrating rational functions?

Some common techniques for integrating rational functions include the method of partial fractions, substitution, and integration by parts. The choice of technique depends on the complexity of the rational function and the skills of the integrator.

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