# Problem resolving an Integral - Partial Fractions

• SclayP
In summary: A + C = 0\ , so C = -A .A + B + C + D = 1\ , so B + D = 1-A\ .Use the value of B you got when you let x = 1 to get D .B + D = 1-A\ , so 1-A = B + D = 0 + D = D\ .D = 1-A\ .Use the value of D you got when you let x = -1 to get A .D = 1-A\ , so 1-A = D = 1-A\ , so 1-A = 1-A\ .A = A
SclayP
1. So, i have the next integrand...
2. $\int \frac{1}{(x-1)^2(x+1)^2}\,dx$
3. I proceeded by resolving it by partial fraction and i came up with the next...

$\int \frac{1}{((x-1)^2)((x+1)^2)}\,dx = \int \frac{A}{(x-1)} + \frac{B}{(x-1)^2} + \frac{C}{(x+1)} + \frac{D}{(x+1)^2}\,dx$

The thing is that after doing all the calculus i came up with this...

$A + C = 0$
$A + B - C + D = 0$
$-A + 2B - C -2D = 0$
$A + B + C + D = 1$

After this i don't know how to preceed i mean i don't know how to resolve the equation whit 4 variables...

Thanks, and very sorry for my english...

Last edited:
SclayP said:
1. So, i have the next integrand...

2. $\int \frac{1}{(x-1)^2(x+1)^2}\,dx$

3. I proceeded by resolving it by partial fraction and i came up with the next...

$\int \frac{1}{((x-1)^2)((x+1)^2)}\,dx = \int \frac{A}{(x-1)} + \frac{B}{(x-1)^2} + \frac{C}{(x+1)} + \frac{D}{(x+1)^2}\,dx$

The thing is that after doing all the calculus i came up with this...

$A + C = 0$
$A + B - C + D = 0$
$-A + 2B - C -2D = 0$
$A + B + C + D = 1$

After this i don't know how to proceed i mean i don't know how to resolve the equation whit 4 variables...

Thanks, and very sorry for my english...
Use $\ \ A + C = 0\ \$ with $\ \ -A + 2B - C -2D = 0\ \$

to get $\ \ 2B -2D = 0\ .$

Similarly, use $\ \ A + C = 0\ \$ with $\ \ A + B + C + D = 1$

to get $\ \ B + D = 1\ .$

Use those two equations to solve for B & D .

Put the results for B & D into the first two equations to get A & C .

As an alternative to the above method, consider the following. (This method was also discussed in the thread: Partial Fractions)

I assume that you got your 4 equations for A, B, C, and D by equating coefficients for powers of x in the following equation.

$1=A(x-1)(x+1)^2+B(x+1)^2+C(x-1)^2)(x+1)+D(x-1)^2\ .$

Your can quickly solve for B, by letting x = 1 .

Similarly, you can solve for D, by letting x = -1 .

After that it's not so difficult to find A & C .

## 1. What is the purpose of using partial fractions to solve integrals?

Partial fractions are used to break down a complex rational function into simpler fractions, making it easier to integrate. This method is particularly useful when dealing with improper fractions or when the denominator cannot be factored using traditional methods.

## 2. How do you determine the coefficients when using partial fractions?

The coefficients of each term in the partial fractions can be found by equating the numerator of the original function to the sum of the numerators of the partial fractions. This can be done by setting up a system of equations and solving for the unknown coefficients.

## 3. Can partial fractions be used to solve all types of integrals?

No, partial fractions can only be used to solve integrals of rational functions, where the denominator can be factored into linear and irreducible quadratic terms.

## 4. Can partial fractions be used to solve improper integrals?

Yes, partial fractions can be used to solve improper integrals, as long as the integrand is a rational function with a proper or improper fraction. However, additional steps may be required to handle any infinite limits of integration.

## 5. Are there any shortcuts or tricks to solving integrals using partial fractions?

Yes, there are several techniques that can make solving integrals using partial fractions easier. These include factoring the denominator, using the method of undetermined coefficients, and using the Heaviside cover-up method. It is recommended to practice these techniques to become more efficient at solving integrals using partial fractions.

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