Partial Fractions Calc Problem

In summary, the given integral can be solved using the partial fraction decomposition method. The solution involves finding the values of A and B in the equation 1=B(a-b), and then using the knowledge of the integral of 1/x to solve for the final answer.
  • #1

Homework Statement


∫1/[(x+a)(x+b)]dx
answer is 1/(a-b) ln[(x+b)/(x+a)] + C


Homework Equations





The Attempt at a Solution



1=A/(x+a) + B/(x+b)
1=B(x+a) + A(x+b)
1=Bx+ Ba + Ax +Ab

so 0=Bx+ Ax, 1=Ba+Ab
A=-B, 1=B(a-b)

∫-1/(x+a) +∫1/(x+b)

Im not sure if I am headed in the right direction or not.
 
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  • #2
coookiemonste said:

The Attempt at a Solution



1=A/(x+a) + B/(x+b)

This line should really be 1/((x+a)(x+b))=A/(x+a) + B/(x+b)

1=B(x+a) + A(x+b)
1=Bx+ Ba + Ax +Ab

so 0=Bx+ Ax, 1=Ba+Ab
A=-B, 1=B(a-b)

Good, so B=___?

∫-1/(x+a) +∫1/(x+b)

Im not sure if I am headed in the right direction or not.

This should be B∫-1/(x+a) +B∫1/(x+b)
 
  • #3
How do you know what A and B equal? You don't know numerical values, just what they're equal to in relation to one another.

gabbagabbahey is right, you should get B∫-1/(x+a) +B∫1/(x+b)

Now what do you know about the integral of 1/x ?
 
  • #4
muso07 said:
How do you know what A and B equal? You don't know numerical values, just what they're equal to in relation to one another.
The equation 1= B(a-b) settles that.

gabbagabbahey is right, you should get B∫-1/(x+a) +B∫1/(x+b)

Now what do you know about the integral of 1/x ?
 

1. What are partial fractions and why are they used in calculus?

Partial fractions are a method used in calculus to simplify and evaluate complex rational expressions. They are used to break down a single fraction into smaller, more manageable fractions that are easier to integrate or manipulate.

2. How do you solve a partial fractions problem?

To solve a partial fractions problem, you first need to factor the denominator into linear or quadratic factors. Then, you set up a system of equations using the unknown coefficients for each factor. Finally, you solve for the coefficients and rewrite the original expression as a sum of the partial fractions.

3. What is the difference between proper and improper partial fractions?

A proper partial fraction has a denominator with a higher degree than the numerator, meaning the fraction is less than one. An improper partial fraction has a numerator with a higher degree than the denominator, making the fraction greater than one.

4. Can you use partial fractions to integrate a rational function?

Yes, partial fractions can be used to integrate rational functions. By breaking down a complex rational function into smaller fractions, integration becomes easier and more manageable. This method is commonly used in calculus to solve integrals involving rational functions.

5. Are there any limitations or restrictions when using partial fractions?

Yes, there are some limitations and restrictions when using partial fractions. The denominator of the original fraction must be factorable into linear or quadratic factors. Additionally, the factors must be distinct and irreducible, meaning they cannot be further simplified. If these conditions are not met, the partial fractions method may not be applicable.

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