Par$al Fac$on Decomposi$on Academic Resource Center
Table of Contents. What is Par$al Frac$on Decomposi$on 2. Finding the Par$al Fac$on Decomposi$on 3. Examples 4. Exercises 5. Integra$on with Par$al Fac$ons 6. Example 7. Exercises 8. Resources
What is Par$al Frac$on Decomposi$on This technique should be used to take a single frac$on and separate it into the sum of simpler frac$ons This is can be used to simplify a term for taking the deriva$ve This also works to simplify a term for integra$on
Finding the Par$al Frac$on Decomposi$on Consider a ra$onal func$on ƒ(x)=p(x)/q(x) where P and Q are polynomials. It s possible to express f as a sum of simpler frac$ons, assuming P has a smaller degree than Q. If P has a larger degree than Q we must first divide P by Q so as to get a remainder R(x), which would produce the following equa$on: ƒ(x)=s(x)+r(x)/q(x)
Example Simplify x3 + x x Sincethetop has a higher power than the bottom we must dolong division first x 2 + x + 2 x x 3 + 0 * x 2 + x + 0 x 3 x 2 x 2 + x x 2 x 2x + 0 2x 2 2 x 3 + x x = x2 + x + 2 + 2 x
Finding the Par$al Frac$on Decomposi$on The next step in par$al frac$on decomposi$on is to factor Q(x) as much as possible. Then express R(x)/Q(x) as a sum of par$al frac$ons of the form A (ax + b) i or Ax + B (ax 2 + bx + c) j These steps can be broken down into 4 cases.
Case I: Q(x) is a product of dis$nct linear factors This means that Q(x) can be rewrizen as the following Q(x) = (a x + b )(a 2 x + b 2 ) (a k x + b k ) This means that R(x)/Q(x) can be rewrizen as the following R(x) Q(x) = A a x + b + A 2 a 2 x + b 2 + + A k a k x + b k
Example 2 Simplify x 2 + 2x 2x 3 + 3x 2 2x First we must factor Q(x) 2x 3 + 3x 2 2x = x(2x 2 + 3x 2) = x(2x )(x + 2) Now our fac$on can be rewrizen as follows x 2 + 2x 2x 3 + 3x 2 2x = A x + B 2x + C x + 2
Example 2 cont. To determine the values of A,B, and C, we mul$ply both sides by x(2x- )(x+2) which results in the following x 2 + 2x = A(2x )+ Bx(x + 2)+ Cx(2x ) A^er expanding and rewri$ng the equa$on we obtain x 2 + 2x = (2A + B C)x 2 + (3A + 2B C)x 2A
Example 2 cont. Since the polynomials are iden$cal, there coefficients must be equal. This gives us the following system of equa$ons 2A + B + 2C = 3A + 2B C = 2 2A = Solving this we get the following equa$on for our original func$on x 2 + 2x 2x 3 + 3x 2 2x = 2 x + 5 2x 0 x + 2
Case II: Q(x) is a product of linear factors, some of which repeat Suppose that Q(x) is the product of one linear factor, r $mes. This means that Q(x) can be rewrizen as the following Q(x) = (a x + b ) r This means that R(x)/Q(x) can be rewrizen as the following R(x) Q(x) = A a x + b + A 2 (a x + b ) 2 + + A r (a x + b ) r
Example 3 Simplify x 4 2x 2 + 4x + x 3 x 2 x + The first step is long division resul$ng in the following x 4 2x 2 + 4x + x 3 x 2 x + = x ++ 4x x 3 x 2 x + The next step is to factor Q(x) giving us the following x 3 x 2 x += (x )(x 2 ) = (x )(x )(x +)
Example 3 cont. Now our R(x)/Q(x) can be rewrizen as follows 4x x 3 x 2 x + = A x + B (x ) + C 2 x + Next we solve for A, B, and C and get our solu$on 4x = A(x )(x +)+ B(x +)+ C(x ) 2 = (A + C)x 2 + (B 2C)x + ( A + B + C) A + C = 0 A = B 2C = 4 B = 2 A + B + C = 0 C = x 4 2x 2 + 4x + x 3 x 2 x + = x ++ x + 2 (x ) 2 x +
Case III: Q(x) contains irreducible quadra$c factors, which do not repeat Suppose that Q(x) has a factor ax 2 +bx+c which can not be factored, then, in addi$on to the par$al factors, R(x)/Q(x) will have a term of the form Ax + B ax 2 + bx + c
Example 4 Simplify 2x2 x + 4 x 3 + 4x Since x 3 +4x=x(x 2 +x) can not be factored, we write 2x 2 x + 4 x(x 2 + 4) = A x + Bx + C x 2 + 4 By mul$plying both sides by x(x 2 +4) we get 2x 2 x + 4 = A(x 2 + 4)+ (Bx + C)x = (A + B)x 2 + Cx + 4A
Example 4 cont. Now solving for A, B, and C we get A + B = 2 C = 4A = A A =, B =, and C = So we see that our func$on reduces to this 2x 2 x + 4 x 3 + 4x = x + x x 2 + 4
Case IV: Q(x) contains a repeated irreducible quadra$c factor Suppose that Q(x) has a factor ax 2 +bx+c which can not be factored and is repeated r $mes, then, in addi$on to the par$al factors, R(x)/Q(x) will have a terms of the form A x + B ax 2 + bx + c + A 2 x + B 2 (ax 2 + bx + c) 2 + + A r x + B r (ax 2 + bx + c) r
Example 5 Simplify x + 2x2 x 3 x(x 2 +) 2 The form of the par$al frac$on decomposi$on is x + 2x 2 x 3 x(x 2 +) 2 = A x + Bx + C x 2 + + Dx + E (x 2 +) 2 By mul$plying both sides by x(x 2 +) 2 we get x 3 + 2x 2 x += A(x 2 +) 2 + (Bx + C)x(x 2 +)+ (Dx + E)x = A(x 4 + 2x 2 +)+ B(x 4 + x 2 )+ C(x 3 + x)+ Dx 2 + Ex = (A + B)x 4 + Cx 3 + (2A + B + D)x 2 + (C + E)x + A
Example 5 cont. Now solving for A, B, C, D, and E we get A + B = 0 C = 2A + B + D = 2 C + E = A = A =, B =,C =, D =, and E = 0 So we see that our func$on reduces to this x + 2x 2 x 3 x(x 2 +) 2 = x x + x 2 + + x (x 2 +) 2
Exercises Find the partial fractions for the following but not thecoefficients. 2. 3. x 3 + 2x 2 + x x 4 + x 5 + 4x 3 t 4 + t 2 + (t 2 +)(t 2 + 4) 2
Answers. 2. 3. A x + B x + + C (x +) 2 A x + B x + C 2 x + Dx + E 3 x 2 + 4 At + B t 2 + + Ct + D t 2 + 4 + Et + F (t 2 + 4) 2
Integra$on with Par$al Frac$ons Par$al frac$on decomposi$on is useful in integra$on when the term that you are integra$ng is a func$on such that can be wrizen as the division of two polynomials
Example 6 Let s try to integrate the func$on in Example 3 x 2 + 2x 2x 3 + 3x 2 2x dx = " $ # 2 = 2 x + 5 x dx + 5 = 2 ln x + 0 2x 0 x + 2 2x dx 0 % 'dx & x + 2 dx ln 2x 0 ln x + 2 + C
Exercises. 2. 3. 3 2 2 x 2 dx 4y 2 7y 2 y(y + 2)(y 3) dy 0 (x )(x 2 + 9) dx
Answers. 2. 2 ln(3 2 ) 27 5 ln(2) 9 5 ln(3) 3. ln x 2 ln(x2 + 9) 3 tan (x / 3)+ C
Resources Calculus, Chapter 8.4 hzp://mathworld.wolfram.com/ Par$alFrac$onDecomposi$on.html