Help Understanding Integral with Partial Fractions

In summary, the conversation involved a question about understanding a certain integral and the steps taken to solve it using partial fractions. The issue arose when checking the answer with Mathematica, as it gave a slightly different result. It was clarified that the difference was due to a constant of integration and that different constants may be needed in different ranges.
  • #1
Airsteve0
83
0
Hey everyone, I was wondering if someone could help me understand what exactly is happening with a certain integral I am working with, which is as follows:

∫2/(u^2-1)du

My steps are as follows (I used partial fractions):

∫(1/(u-1) - 1/(u+1))du = ∫1/(u-1)du - ∫1/(u+1) = ln[(u-1)/(u+1)]

However, here is where my issue arrises; when checking my answer with Mathematica, if I input the very first line above I get:

ln[(1-u)/(1+u)]

Could someone help me understand if it is my method that is flawed or maybe the way I am inputting it into the program when I check my answer. Any assistance is greatly appreciated, thanks!
 
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  • #2
ln[(u-1)/(u+1)] = ln[(-1 +u)/(1+u)] = ln[(-1)(1-u)/(1+u)] =ln(-1) + ln[(1-u)/(1+u)] .

Since the integral is indefinite, there must be a constant of integration. The ln(-1) can be incorporated in this constant.

So if you add a constant to your solution, your answer and that given by Matematica will be equivalent.
 
  • #3
Both answers are correct as they differ by a constant.
 
  • #4
o ok so the constant can be complex in general then?
 
  • #5
Yes the constant would be complex. To avoid dealing with this often absolute values are used since otherwise the function would be undefined. In fact different constants are needed in different ranges.
 
  • #6
thank you very much for the clarification and assistance!
 

Related to Help Understanding Integral with Partial Fractions

1. What is the purpose of using partial fractions in integrals?

Partial fractions are used to simplify complex rational functions into smaller, more manageable parts. This allows for easier integration and evaluation of the function.

2. How do I know when to use partial fractions in an integral?

Partial fractions are typically used when the integrand is a rational function (a polynomial divided by another polynomial). Additionally, if the degree of the numerator is greater than or equal to the degree of the denominator, partial fractions can be used.

3. What are the steps for solving an integral using partial fractions?

The general steps for solving an integral using partial fractions are:

  1. Factor the denominator of the rational function into linear or irreducible quadratic factors.
  2. Set up a system of equations using the coefficients of the partial fractions.
  3. Solve the system of equations to find the values of the coefficients.
  4. Rewrite the original integral using the partial fraction decomposition.
  5. Integrate each term separately.
  6. Combine the integrals and simplify to get the final answer.

4. What is the difference between partial fractions with repeated factors and distinct factors?

Partial fractions with repeated factors occur when the denominator of the rational function has the same factor more than once. In this case, the corresponding partial fraction must be multiplied by a power of x. Distinct factors, on the other hand, occur when the denominator has different factors, each with a unique coefficient in the partial fraction decomposition.

5. Can I use partial fractions to solve any integral?

No, partial fractions can only be used for integrals involving rational functions. If the integrand is not a rational function, other integration techniques such as substitution or integration by parts may be necessary.

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