1 Integration of Rational Functions Using Partial Fractions

Transcription

1 MTH Fall 008 Essex County College Division of Mathematics Handout Version 4 September 8, 008 Integration of Rational Functions Using Partial Fractions In the past it was far more usual to simplify or combine fractions than it was to expand them. However, it is sometimes easier to break an integrand up into simpler problems, and for rational expressions we have several forms to recognize. nd I want to again emphasize that recognition does not necessarily mean memorization. Here s a partial list that should be recognized, and I am putting a boxes around those that may be helpful when integrating rational functions:. sinh x ) =. Which of course leads to: + x + x = sinh x + C cosh x ) =. Which of course leads to: x x = cosh x + C tanh x ) =. Which of course leads to: x x = tanh x + C csch x ) = x. Which of course leads to: + x x + x = csch x + C sech x ) = x. Which of course leads to: x x x = sech x + C coth x ) =. Which of course leads to: x x = coth x + C This document was prepared by Ron Bannon using L TEX ε.

2 I think, however, that these should probably be memorized, at least for an exam.. x = arcsin x + C.. = arctan x + C + x x = ln x + C The Pre-Calulus of Partial Fraction Decomposition You ll have a rational function in lowest terms of the form Then you ll need to follow the steps below. P (x) Q (x).. If the degree of the numerator is greater than or equal to the denominator, long divide. You will be left with a rational function. For example: 6x + x 7 x x = x + + 8x 4 x x.. If the degree of the numerator is less than the degree of the denominator, then factor the denominator into linear factors or irreducible quadratic factors. n irreducible quadratic factor is a quadratic that cannot not be factored into linear factors with real coefficients. Example continued from above. 6x + x 7 x x = x + + 8x 4 (x + ) (x ).. ssign each linear factor (px + q) n the sum of n partial fractions: (px + q) + (px + q) + (px + q) + + n (px + q) n Continuing with the example above. 6x + x 7 x x = x + + 8x 4 (x + ) (x ) = x + + k (x + ) + k (x ) Of course we will need to determine the values of the constants k and k. quadratic polynomial ax + bx + c is called irreducible if it cannot be factored into linear factors without using complex numbers. e.g. x + = (x i) (x + i)

3 4. ssign each irreducible quadratic factor ( ax + bx + c ) m the sum of m partial fractions: B x + C (ax + bx + c) + B x + C (ax + bx + c) + B x + C (ax + bx + c) + + B mx + C m (ax + bx + c) m. gain you ll need to determine these constants. This looks nightmarish, I know, but the algebra is really quite easy. Of course, let s look at some examples.. Preliminary Examples. Decompose into partial fractions. 4x x + x 6 Work: The first step is to break the fraction into two parts. 4x x + x 6 = x + + Now add the fractions on the right hand side together. B x 4x (x ) + B (x + ) x = + x 6 x + x 6 Now equate the numerators, we really no longer need to worry about the denominator, after-all they re the same. Basically we just need to find the constants that make these numerators equal. You should, of course, pick values for x that make solving for and B easy. Let s see what values for x make our job easier. 4x = (x ) + B (x + ) I say we choose x = to see what happens. 4x = (x ) + B (x + ) 8 = ( 4 ) + B ( + ) = 7 = Okay, now that we know that =, let s find B by choosing an easy value for x, let s try x = 0, but any value for x will do just fine. 4x = (x ) + B (x + ) = ( ) + B () = + B 4 = B = B Now we finally have it. 4x x + x 6 = x + x

4 Why bother though? Look at these equivalent integrals and see why! 4x x + x 6 = x + x = x + x Of course we can now integrate. x + x = ln x + ln x + C. Let s return to the first example given, but this time I want to integrate. 6x + x 7 x x = 8x 4 x + + x x = x + + (x + ) + B (x ) Work: Now let s determine the constants and B. 8x 4 = (x ) + B (x + ) Let x = and you ll get: Now let x = 0 and you ll get: 4 = 4B, B =. 4 = +, =. The we have: 6x + x 7 x x = = x + + x + + 8x 4 x x (x + ) + (x ) Finally we need to do the simple integration: x + + (x + ) + (x ) = x + x + ln x + + ln x + C. We have, from page one, and a prior work sheet: x = tanh x + C. Try to integrate this one using partial fraction decomposition. 4

5 Work: x = = = = x + B + x x + B + x / x + / + x x + We ll do this in class. + x = ln x + ln + x + C = ln + x x + C You should not that we have troubles when x = ±, just look at the original integrand. Furthermore if x (, ) we have ln + x x = ( ) + x ln = tanh x x Let s take a look at a definite integral. Figure : Integration of interest. For example, this makes no sense because x / (, ): but this does: x = tanh x ] = tanh tanh = calculator may fail here, x = ln + x x ] = ln ln = ln 0.07 Take home message: start to use technology and have a good book (no, not the Bible) to consult. ctually, each of these terms gives a complex number: tanh tanh = ln π i ln π r i = ln

6 . Examples. Evaluate x + (x ) (x 8) (x + ) x +. Evaluate x 4. Evaluate x (x ) (x + ) 4. Evaluate 8 (x + ) (x + ). Evaluate 4 x x (x + ) 6. Evaluate x + x Partial fractions, as you now know, can be quite laborious. So please consider using a computer algebra system to do some of the above problems. See next page for the Mathematica code for these problems. Greek To Me! Let s be thankful that we have a native Greek speakers (Yes, there here at ECC) to help us out when we get stuck. Lower case greek letter used in mathematics. alpha = α, beta = β, gamma = γ, delta = δ, epsilon = ɛ, var epsilon = ε, zeta = ζ, eta = η =, theta = θ, var theta = ϑ, iota = ι, kappa = κ, lambda = λ, mu = µ, nu = ν, xi = ξ, pi = π, var pi = ϖ, rho = ρ, var rho = ϱ, sigma = σ, var sigma = ς, tau = τ, upsilon = υ, phi = φ, var phi = ϕ, chi = χ, psi = ψ, omega = ω. Upper case Greek letters used in mathematics: gamma = Γ, delta =, theta = Θ, lambda = Λ, xi = Ξ, pi = Π, sigma = Σ, upsilon = Υ, phi = Φ, psi = Ψ, Omega = Ω. 6

7 Mathematica Code.nb In[8]:= + L ê HHx - L H x - 8L Hx + L L, xd Out[8]= In[]:= Out[]= In[0]:= ÅÅÅÅ JLog@-4 + xd - ÅÅÅÅ Log@- + xd + ÅÅÅÅ Log@ + xdn Integrate@Hx^ + L ê Hx^ - 4L, xd x ÅÅÅÅÅÅ + ÅÅÅÅ 4 7 Log@- + xd + ÅÅÅÅ Log@ + xd 4 Integrate@H x - L ê HHx - L Hx + L^ L, xd Out[0]= J- ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅ H + xl - ÅÅÅÅ Log@- + xd + ÅÅÅÅ Log@ + xdn In[]:= Integrate@H8L ê HHx + L Hx^ + L L, xd Out[]= 8 J ÅÅÅÅÅÅÅ 8 rctan ÅÅÅÅ x E + ÅÅÅÅÅÅÅ 8 Log@ + xd - ÅÅÅÅÅÅÅ 6 Log@ + x DN In[]:= Out[]= In[]:= Out[]= In[4]:= Out[4]= In[]:= part@hx^ - L ê HHx + L Hx^ + 4L^LD H- + xl ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ + ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅ 6 H + xl 4 H4 + x L + - x ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ H4 + x L + - x ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅ 6 H4 + x L part@hx^ + L ê HHx - L H x - 8L Hx + L LD ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ H-4 + xl - ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅ 6 H- + xl + ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅ 6 H + xl part@hx^ + L ê Hx^ - 4LD ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ 4 H- + xl + x + 7 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅ 4 H + xl part@h x - L ê HHx - L Hx + L^ LD Out[]= - ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅ H- + xl + ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ H + xl + ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅ H + xl In[6]:= Out[6]= In[7]:= Out[7]= In[8]:= Out[8]= part@h8l ê HHx + L Hx^ + L LD ÅÅÅÅÅÅÅÅÅÅÅ + x + ÅÅÅÅÅÅÅÅÅÅÅÅÅ - x + x part@hx^ - L ê HHx + L Hx^ + 4L^LD H- + xl ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ + ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅ 6 H + xl 4 H4 + x L + - x ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ H4 + x L + - x ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅ 6 H4 + x L part@ ê Hx^ + x + LD ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ + x + x Figure : Mathematica Code 7