Integration+by+Partial+Fractions+(BC)

Integration by partial fractions is a means of finding the integral of a fractional, rational function. This is done by expanding the fraction into a sum of smaller fractions that are equivalent to the original, called partial fraction decomposition. By definition, a partial fraction is any one of a set of simpler fractions, the sum of which composes a more complex fraction. For instance, the partial fraction decompostion for 2/3 would be 1/3 + 1/3. This can be done with simple as well as more complex fractions, generally denoted To demonstrate the method of partial fraction decompostion, let's begin with the fraction 1. Factor the denominator.

2. For the partial fractions, assumer that there are unknown constant variables, and present them with variables A and B, as shown. //*Note: It can be assumed that// A //and// B //are constants and exist for any rational function if the following conditions are met:// //If this condition is not met, that is, if the degree of// p(x) //is equal to or greater than// q(x)//, then polynomial division must be used to rewrite the fraction as the sum of a polynomial and a new rational function that satisfies Condition 2. See link at end of page for further elaboration on polynomial division.// 3. Find a common denominator for the new fractions found in Step 2. Therefore,
 * 1) Both //p(x)// and //q(x)//are polynomials
 * 2) The degree of the numberator //p(x)// is smaller than the degree of the denominator //q(x)//.

4. Having common denominators, the numerators can now be set equal.

5. Since this equation is true for all x values, choose an x to input and solve for both //A// and //B.// (In this case, x=2 and x=-2 can be used to "zero out" either variable and solve for the other.) Once solved, A=-3/4 and B=3/4.

6. Thus, OR To find the integral of this partial fraction decomposition, use the new set of fractions as the integrand and take the integral (using u-subsititution, or any other applicable method) to solve as usual.

7.

//**Practice Problems**//

//**Multiple Choice:**//

1.

A. B.  C. D.  E.

2. Which of the following is the solution to the initial value problem and ?

A. B. C.  D. E.



//**Free Response:**//

1. Integrate.



Link for polynomial division: []

Sources Used: [] [] []