Calculus II. Monday, March 13th. WebAssign 7 due Friday March 17 Problem Set 6 due Wednesday March 15 Midterm 2 is Monday March 20

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1 Announcements Calculus II Monday, March 13th WebAssign 7 due Friday March 17 Problem Set 6 due Wednesday March 15 Midterm 2 is Monday March 20 Today: Sec. 8.5: Partial Fractions Use partial fractions to decompose a rational fraction Integrate using partial fractions and other techniques Next Class: Sec. 8.8: Improper Integrals

2 Intro to Partial Fractions Goal: Calculate N(x) dx where N, D are polynomials. D(x)

3 Intro to Partial Fractions Goal: Calculate N(x) dx where N, D are polynomials. D(x) If deg(n) deg(d), can do long division.

4 Intro to Partial Fractions Goal: Calculate N(x) dx where N, D are polynomials. D(x) If deg(n) deg(d), can do long division. Can assume deg(n) < deg(d)

5 Intro to Partial Fractions Goal: Calculate N(x) dx where N, D are polynomials. D(x) If deg(n) deg(d), can do long division. Can assume deg(n) < deg(d) Fundamental Theorem of Algebra

6 Intro to Partial Fractions Goal: Calculate N(x) dx where N, D are polynomials. D(x) If deg(n) deg(d), can do long division. Can assume deg(n) < deg(d) Fundamental Theorem of Algebra Every polynomial is a product of linear factors over C.

7 Intro to Partial Fractions Goal: Calculate N(x) dx where N, D are polynomials. D(x) If deg(n) deg(d), can do long division. Can assume deg(n) < deg(d) Fundamental Theorem of Algebra Every polynomial is a product of linear factors over C. - Not every real polynomial factors completely over R...

8 Intro to Partial Fractions Goal: Calculate N(x) dx where N, D are polynomials. D(x) If deg(n) deg(d), can do long division. Can assume deg(n) < deg(d) Fundamental Theorem of Algebra Every polynomial is a product of linear factors over C. - Not every real polynomial factors completely over R but complex roots must come in conjugate pairs

9 Intro to Partial Fractions Goal: Calculate N(x) dx where N, D are polynomials. D(x) If deg(n) deg(d), can do long division. Can assume deg(n) < deg(d) Fundamental Theorem of Algebra Every polynomial is a product of linear factors over C. - Not every real polynomial factors completely over R but complex roots must come in conjugate pairs - A product of conjugate pair factors is a real quadratic

10 Intro to Partial Fractions Goal: Calculate N(x) dx where N, D are polynomials. D(x) If deg(n) deg(d), can do long division. Can assume deg(n) < deg(d) Fundamental Theorem of Algebra Every polynomial is a product of linear factors over C. - Not every real polynomial factors completely over R but complex roots must come in conjugate pairs - A product of conjugate pair factors is a real quadratic Can assume D(x) is a product of degree 1 and 2 factors

11 Partial Fractions: Distinct Linear Factors Theorem If deg(n) < deg(d) and D(x) is a product of distinct linear factors then D(x) = L 1 (x) L 2 (x) L 3 (x) N(x) D(x) = A L 1 (x) + for some constants A, B, C,... B L 2 (x) + C L 3 (x) +

12 Partial Fractions: Distinct Linear Factors Theorem If deg(n) < deg(d) and D(x) is a product of distinct linear factors then D(x) = L 1 (x) L 2 (x) L 3 (x) N(x) D(x) = A L 1 (x) + for some constants A, B, C,... B L 2 (x) + C L 3 (x) + Finding Constants: Combine the right side into one fraction and set the resulting numerator equal to N(x).

13 Partial Fractions: General Case Theorem If deg(n) < deg(d), then N(x) D(x) is a sum of terms of the form: For each linear factor L(x) of D(x) with multiplicity N: A L(x) + B L(x) N L(x) N For each irreducible quadratic factor Q(x) of D(x) with multiplicity M: Ax + B Q(x) + Cx + D Q(x) Yx + Z Q(x) M

14 Partial Fractions: General Case Theorem If deg(n) < deg(d), then N(x) D(x) is a sum of terms of the form: For each linear factor L(x) of D(x) with multiplicity N: A L(x) + B L(x) N L(x) N For each irreducible quadratic factor Q(x) of D(x) with multiplicity M: Ax + B Q(x) + Cx + D Q(x) Yx + Z Q(x) M Note: Constants got reused here for notational simplicity. Use each unknown constant only once in a problem.

15 Practice Example What s the form of the partial fraction decomposition? (a) 2x 3 1 (x + 1) 3 (x 2 + 4x + 7) (b) 3x + 2 x 5 + 4x 3 + 4x [Don t solve for the constants!]

16 Practice Example What s the form of the partial fraction decomposition? (a) 2x 3 1 (x + 1) 3 (x 2 + 4x + 7) (b) 3x + 2 x 5 + 4x 3 + 4x [Don t solve for the constants!] Answers: (a) x 2 + 4x + 7 is irreducible (quadratic formula), (b) 2x 3 1 (x + 1) 3 (x 2 + 4x + 7) = A x B (x + 1) 2 + C (x + 1) 3 + Dx + E x 2 + 4x + 7 3x + 2 x 5 + 4x 3 + 4x = A x + Bx + C x Dx + E (x 2 + 2) 2

17 Example Calculate x x (x 1)(x + 1) 2 dx.

18 Example Calculate x x (x 1)(x + 1) 2 dx. Answer: x x (x 1)(x + 1) 2 dx = 3 x x (x + 1) 2 dx = 3 ln x 1 2 ln x x C

19 Example Calculate x x 4 dx.

20 Example Calculate x x 4 dx. Answer: Let u = x. x x 4 dx = 2u 2 u 2 4 du = u 2 1 u + 2 du = 2u + 2 ln u 2 u C = 2 x + 2 ln x 2 + C x + 2

21 Mixed Practice 1. What technique would you use to integrate? x + 1 (a) x 2 + 2x 8 dx 7x + 4 (b) x 2 + 2x 8 dx 4 (c) x 2 + 2x + 5 dx 2. Use your integration techniques to find: sec 2 x (a) tan x(tan x + 1) dx (b) ln(x 2 + 1) dx x 1 (c) (Very long) x(x 2 + 1) 2 dx

22 Mixed Practice Answers 1. (a) u-sub (b) Partial fractions (c) Trig sub 2. (a) With u = tan x, 1 1 u(u + 1) du = u 1 u + 1 du = = ln tan x tan x C (b) Do integration by parts first, then long division. ln(x 2 + 1) dx = = x ln(x 2 + 1) 2x + 2 tan 1 (x) + C (c) Partial fractions. Most resulting terms use u-sub and one needs trig sub. Final answer is ln x + 1 [ ln x (x 2 + 1) + tan 1 x + x ] x 2 + C + 1