# Integration by partial fractions with limits

• anthonyk2013
In summary, the integral given is (x + 1)/(x2 - 3x - 5) and the limits of integration are 3 and 5. The integrand should be written in parentheses and the integral can be solved using partial fractions. The limits of integration can be factored in later in the solution. However, it may be easier to solve the integral without using partial fractions due to the messy nature of the denominator.
anthonyk2013
$\int$ (x+1/x2-3x-5)dx

I can't put the limits on the integral sign, 5 is the top limit and 3 is the bottom limit.

I can solve using partial fractions ok but I have never solved with limits before.

Where do the limits come in, do I need them at the start or can I factorise as usual and use the limits later in the solution?

anthonyk2013 said:
$\int$ (x+1/x2-3x-5)dx

I can't put the limits on the integral sign, 5 is the top limit and 3 is the bottom limit.

I can solve using partial fractions ok but I have never solved with limits before.

Where do the limits come in, do I need them at the start or can I factorise as usual and use the limits later in the solution?

You have used parentheses, but they're not where they need to be. When a numerator or denominator consists of multiple terms, put parentheses around the entire numerator and/or the entire denominator. I gather that the integrand is (x + 1)/(x2 - 3x - 5). As you wrote the integrand, it would be the same as x + (1/x2) - 3x - 5, which isn't what you intended.

To write the integral with limits in LaTeX, you can do it like this:
Code:
$$\int_3^5 \frac{x + 1}{x^2 - 3x - 5}dx$$

I prefer to use $$instead of [ tex ]. For inline LaTeX, you can use ## instead of [ itex ]. To answer your question, all you're doing when you use partial fractions is rewriting the integrand in a more convenient form. You bring the limits of integration along until you actually get the antiderivative. For this problem, I probably wouldn't use partial fractions, since the denominator isn't going to factor into nice linear factors with integer coefficients. I'm not saying it won't work, but it will be a little messy. Mark44 said: You have used parentheses, but they're not where they need to be. When a numerator or denominator consists of multiple terms, put parentheses around the entire numerator and/or the entire denominator. I gather that the integrand is (x + 1)/(x2 - 3x - 5). As you wrote the integrand, it would be the same as x + (1/x2) - 3x - 5, which isn't what you intended. To write the integral with limits in LaTeX, you can do it like this: Code: $$ \int_3^5 \frac{x + 1}{x^2 - 3x - 5}dx$$ I prefer to use$$ instead of [ tex ]. For inline LaTeX, you can use ## instead of [ itex ].

To answer your question, all you're doing when you use partial fractions is rewriting the integrand in a more convenient form. You bring the limits of integration along until you actually get the antiderivative.

For this problem, I probably wouldn't use partial fractions, since the denominator isn't going to factor into nice linear factors with integer coefficients. I'm not saying it won't work, but it will be a little messy.

Thanks

1.

## What is integration by partial fractions with limits?

Integration by partial fractions with limits is a method used in calculus to integrate a rational function, which is a fraction where the numerator and denominator are both polynomial functions. It involves breaking down the rational function into smaller, simpler fractions and then integrating each term individually.

2.

## When is integration by partial fractions with limits used?

This method is often used when the original function is difficult to integrate using other methods, such as substitution or integration by parts. It is also useful when dealing with improper integrals, which have limits that approach infinity or negative infinity.

3.

## How do you find the partial fractions of a rational function?

To find the partial fractions of a rational function, the numerator and denominator must first be factored completely. Then, each distinct linear factor in the denominator will have its own fraction in the partial fraction decomposition. If there are repeated factors, the decomposition will include a fraction for each power of that factor.

4.

## What are the steps for integrating a rational function using partial fractions with limits?

The steps for integrating a rational function using partial fractions are as follows:
1. Factor the numerator and denominator of the rational function.
2. Write the partial fraction decomposition, with a fraction for each distinct linear factor in the denominator.
3. Set up equations to solve for the unknown coefficients in each fraction.
4. Use algebra to solve for the coefficients.
5. Integrate each term in the partial fraction decomposition.
6. Substitute the limits of integration into the integrated terms and evaluate the resulting expressions.
7. Add the evaluated expressions together to find the final result of the integration.

5.

## What are some common mistakes when using integration by partial fractions with limits?

Common mistakes when using integration by partial fractions include:
- Forgetting to factor the numerator and denominator of the rational function before writing the partial fraction decomposition.
- Not setting up the correct equations to solve for the unknown coefficients.
- Making errors when solving for the coefficients.
- Forgetting to integrate each term in the partial fraction decomposition.
- Making mistakes when substituting the limits of integration into the integrated terms.
It is important to carefully follow each step and double check for errors to ensure an accurate result.

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