Partial Fraction Decomposition Help - Calculus BC

In summary, we are asked to decompose and integrate a rational function, using the Heaviside cover-up method to find the decomposition coefficients.
  • #1
MarkFL
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Here is the question:

Help with Calculus BC: partial fractions!?

Thank you for your help guys. I wrote the problem wrong in the previous question :(

integral (4x^2+x-2)/(x^3+2x^2-3x) dx

howwwww?

Here is a link to the question:

Help with Calculus BC: partial fractions!? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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  • #2
Hello living higher:

We are given to decompose:

$\displaystyle \frac{4x^2+x-2}{x^3+2x^2-3x}$

The first step is to factorize the denominator:

$x^3+2x^2-3x=x(x^2+2x-3)=x(x-1)(x+3)$

Now we assume the decomposition will take the form:

$\displaystyle \frac{4x^2+x-2}{x(x-1)(x+3)}=\frac{A}{x+3}+\frac{B}{x}+\frac{C}{x-1}$

Now, rather than set up a linear system of equations, I am going to use a shortcut method called the Heaviside cover-up method. Look at the first term on the right. The root of the denominator is $x=-3$.

To find the value of $A$, we "cover-up" the factor $x+3$ to get:

$\displaystyle \frac{4x^2+x-2}{x(x-1)}$

and we evaluate this at $x=-3$ to find:

$\displaystyle A=\frac{4(-3)^2+(-3)-2}{(-3)((-3)-1)}=\frac{31}{12}$

Next we look at the second term in the decomposition and we see the root of the denominator is $x=0$, and covering up the factor $x$ on the left, and evaluating it for $x=0$, we find:

$\displaystyle B=\frac{4(0)^2+(0)-2}{((0)-1)((0)+3)}=\frac{2}{3}$

And finally, we look at the third term in the decomposition and we see the root of the denominator is $x=1$, and covering the the factor $x-1$ on the left and evaluation it for $x=1$, we find:

$\displaystyle C=\frac{4(1)^2+(1)-2}{(1)((1)+3)}=\frac{3}{4}$

And now we may state:

$\displaystyle \frac{4x^2+x-2}{x^3+2x^2-3x}=\frac{31}{12(x+3)}+\frac{2}{3x}+\frac{3}{4(x-1)}$

Now we may directly integrate:

$\displaystyle \int\frac{31}{12(x+3)}+\frac{2}{3x}+\frac{3}{4(x-1)}\,dx=\frac{31}{12}\ln|x+3|+\frac{2}{3}\ln|x|+ \frac{3}{4}\ln|x-1|+C$
 

FAQ: Partial Fraction Decomposition Help - Calculus BC

What is Partial Fraction Decomposition?

Partial Fraction Decomposition is a method used in calculus to break down a rational function into simpler fractions. It is especially helpful when integrating rational functions, as it allows for easier integration.

Why is Partial Fraction Decomposition useful?

Partial Fraction Decomposition is useful because it simplifies complicated rational functions, making them easier to integrate. It also allows for the use of other integration techniques, such as u-substitution, to solve the integral.

How do I perform Partial Fraction Decomposition?

To perform Partial Fraction Decomposition, you must first factor the denominator of the rational function. Then, you must set up an equation with unknown coefficients for each unique factor in the denominator. By solving this system of equations, you can determine the values of the coefficients and write the function as a sum of simpler fractions.

What types of rational functions can be decomposed using Partial Fraction Decomposition?

Partial Fraction Decomposition can be used on any proper rational function, which is a function with a polynomial in the numerator and denominator where the degree of the numerator is less than the degree of the denominator.

Are there any limitations to Partial Fraction Decomposition?

Yes, Partial Fraction Decomposition can only be used on proper rational functions. It also cannot be used on irrational or transcendental functions, such as square roots or trigonometric functions.

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