Partial Fractions. June 27, In this section, we will learn to integrate another class of functions: the rational functions.

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1 Partial Fractions June 7, 04 In this section, we will learn to integrate another class of functions: the rational functions. Definition. A rational function is a fraction of two polynomials. For example, x + x, x 7 + x 5 4, x + x, and 0 are all rational functions (a polynomial is a rational function because it can be written as a fraction with denominator. So how do we approach such a general problem as integrating any rational function? Well, let s start with writing down a rational function as f(x/g(x, where f(x and g(x are polynomials. If the degree of f(x is greater than or equal to that of g(x, then we can divide the polynomials to get f(x g(x = q(x + r(x g(x where q(x is the quotient and r(x is the remainder. We can easily integrate the quotient, so the real question then becomes, how do we integrate a rational function whose numerator has degree less than the denominator, namely r(x/g(x? Let s start off with integrals we know we can do and go from there. We know, for example, = ln x a + C x a for any real number a, and (x a n+ = + C (x a n n + for positive integers n >. So if we can make it so that, somehow, all we have to do is solve such integrals, we are in good shape. In fact, if we could convert any rational function into a sum of such fractions, then we could easily perform the integration.

2 How can we convert a rational function into a sum of fractions where the denominator is a power of a linear term like x a? Well, let s start off by factoring the denominator. Can we factor any denominator? Not always. There is no real procedure for factoring a polynomial that is of degree 5 or more. You can approximate the roots of the polynomial and so approximately factor it, but you are not always going to be able to get exact expressions for the roots. If you take an advanced course in abstract algebra, you will learn exactly why this is (it s called Galois theory. But we don t need to worry about such complexities here. In our class, we will always be able to factor the denominator. You should be familiar with factoring a quadratic, for instance. Thus suppose we are given a denominator that is factored. The denominator could still have irreducible quadratic factors like x + (which cannot be factored over the real numbers, but can be factored over the complex numbers, but we will deal with those later. Let s just take it one step at a time. So first we re going to suppose that we are given a denominator that has only linear factors like x, (x, (x +, (x, etc. Let s do an example. You can easily check that and x( = x x(x = ( x. x But how did we get that? The process is called decomposition into partial fractions. We know what we want, so we simply need to set up the right equations to get it. Namely, for the first example, we wanted to break up the fraction into some multiple (say A of /x plus some multiple (say B of /(. So let s set up the equation: x( = A x + B We just need to solve this equation for A and B. So let s multiply both sides by x(. In general, you will always multiply by the denominator of your original rational function (in this case, the rational function is /(x(. We get = A( + Bx

3 after canceling the polynomials that appeared in the numerator and denominator of the fractions. Now let s break up the right side according to the powers of x. First distribute: = Ax + A + Bx Now combine the coefficients of x and the constant coefficients (there is only one of the latter: = (A + Bx + A This is an equation between two polynomials (remember, A and B are just constants. For the equation to hold, the corresponding coefficients of the polynomials must be equal. The constant coefficient of the polynomial on the left is, and the constant coefficient of the polynomial on the right is A, so A must be! Next, the coefficient of x in the polynomial on the left is 0 (it s not there!, and the coefficient of x in the polynomial on the right is A + B, so A + B must be 0. Thus we have a system of equations for A and B: = A 0 = A + B But A =, so B = A =. That s it! We ve done it! We ve solved for A and B! A =, B = This is precisely how we got the decomposition x( = x above. Now try and do the same process for /(x(x to get x(x = ( x. x There is also a very useful trick for finding A and B without solving a system of equations. For example, let s find the partial decomposition of (x (x +.

4 We start by setting up the equation as we did before: (x (x + = A x + B x + and multiply both sides by (x (x + : = A(x + + B(x. Now, this is an equation that involves polynomials, so equality must hold for all x. That means if we plug in a particular value of x, we will still have equality. Let s plug in x = : + = A( + Solving, we get A = /. Now let s plug in x = : Thus we get B = /, so + = B( (x (x + = (x + (x +. The trick is to plug in the right values for x, precisely the ones that make the different factors of the denominator zero (x = makes x zero and x = makes x + zero. Let s do a more complicated example. Let s decompose x + (x + 4x 5(x + 4x + 4 into partial fractions. First we have to factor the denominator: (x + 4x 5(x + 4x + 4 = (x + 5(x (x +. The linear factor x + appears with power, so the right form of the partial fraction decomposition is x + (x + 5(x (x + = A x B x + C x + + D (x +. (

5 In fact, whenever we have a linear term appearing with power greater than, say (x a n, we always need fractions with denominators x a, (x a, (x a, and so on, up to (x a n. Again, let s multiply both sides of ( by the denominator (x + 5(x (x + : x + = A(x (x+ +B(x+5(x+ +C(x+5(x (x++d(x+5(x First we want to use the trick we just learned of plugging in the right x- values as much as we can, and then we can deduce the rest of the constants from setting up a system of equations as in the beginning. First let s plug in x = : + = B( + 5( + gives B = /54 = /7. Next, let s plug in x = 5: ( 5 + = A( 5 ( 5 + gives A = 4/( 54 = 6/7. Lastly, we plug in x = : ( + = D( + 5(, which gives D = 7/ 9 = 7/9. Finally, we need to find C. To do that, we first have to multiply out the polynomials on the right: x + = A(x +x 4+B(x +9x +4x+0+C(x +6x +x 0+D(x +4x 5 In fact we just need to set up one equation; it just needs to have C in it, so that we can solve for C. For example, let s look at the coefficients of x on both sides. They have to be equal in order for the two polynomials to be equal, so we have = A + B + C = C Thus C = 4 x + (x + 5(x (x + = 6 7(x (x 4 (x (x + 5

6 so (x + (x + 5(x (x + ( 6 = 7(x (x 4 (x (x + = 6 7 ln x ln x 4 ln x (x + + C Don t worry, the numbers usually won t come out this bad, but you do have to be really careful and make sure to get them all right. Here is another example of the setup that you would use for the partial fraction decomposition: = A x + x 4 + 4x (x (x + (x + 4 B (x + C x + + D (x + + E (x + + F x + 4 The point here is just for you to understand how to set up the partial fractions (those fractions on the right side of the equation; let s not bother actually solving for A, B, C, D, E, F this time. Now, what if we have irreducible quadratics like x + 4, x +, or x x + 5? And how do you even know a quadratic is irreducible in the first place? Well, ax + bx + c is irreducible if b 4ac < 0. The point is, we can t solve the equation ax + bx + c = 0 if b 4ac < 0 because the quadratic formula x = b ± b 4ac a does not work: we can t take the square root of a negative number. Please check that x + 4, x +, and x x + 5 are indeed irreducible. At this point, you can solve all the homework problems that don t involve irreducible quadratics, so you might want to do that and then go back and read the rest of this worksheet. If we do have an irreducible quadratic, we put it in the denominator of the fraction and put something like Ax + B in the numerator. Here are a couple of examples of partial fraction decomposition setups: x(x + = A x + Bx + C x + 6

7 x 4x x (x + (x 4x + 9 = A x + B x + Cx + D x + + Ex + F x 4x + 9 Again, please check that x + and x 4x + 9 are indeed irreducible. If we have the irreducible quadratic appearing to a power larger than one, then we do the same thing for it as we did for linear factors: if we have, for example, (x + n, then we use Ax + B x + + Cx + D (x + + Ex + F (x P x + Q (x + n Here s the final example of a setup of a partial fraction decomposition, just to get the point across: x (x (x + (x + x + 4 = A x + B x + C x + + Ex + F x + + Gx + H x + x D x Ix + J (x + x Kx + L (x + x + 4 We would then solve for all the coefficients as we did before, first multiplying both sides by the denominator x (x (x + (x + x + 4. We can plug in x = 0 and x =, which would let us solve for C and D, respectively, but we would have to solve for the rest by setting up an enormous system of equations as before. Of course, something this complicated would never be an actual problem for you to solve. Let s do an easier problem: x x + 4 x(x + 4 = A x + Bx + C x + 4 Multiply both sides by x(x + 4: Now we can plug in x = 0: x x + 4 = A(x (Bx + C(x ( = A(0 + 4 gives us A =. We have to solve for B and C by setting up the system of equations again, so let s do that: x x + 4 = Ax + 4A + Bx + Cx 7

8 x x + 4 = (A + Bx + Cx + 4A Here we have combined the coefficients of x and written the coefficients in decreasing order of degree, i.e. the coefficient of x, then the coefficient of x, and then the constant coefficient. That makes it easy to compare the coefficients of the two polynomials. Now, comparing them, we get three equations - one for the coefficient of x, one for the coefficient of x, and one for the constant coefficient (which I will label because = x 0 : x : = A + B x : = C : 4 = 4A In this case, it turns out we didn t need to plug in x = 0 because it is obvious from the equations that A =. We also see from the second equation that C =. Plugging in A = into the first equation, we get = + B so B =. Thus x x + 4 = x(x + 4 x + x x + 4 Now let s integrate this: (x x + 4 x(x + 4 The second integral, = (x x + x + 4 (x, x + 4 is the hard part. We have to split this integral up into parts using the following logic: we will use the derivative of x + 4, namely x, to make a u-substitution, and work with the rest some other way. (x x = x + 4 x + 4 x + 4 OK, now you get to work a bit on your own: show that doing the u- substitution gives x x + 4 = ln x C 8

9 and show that using the substitution we learned in the last section, namely x = tan θ, gives x + 4 = ( x tan + C In total, (x x + 4 = ln x + ln x + 4 ( x x(x + 4 tan + C The only thing really left to understand is how to solve an integral like (6x + 7 (x + x + 4 Here we are doing an example that is harder than you will see in practice, but if you can understand a harder example, then you will definitely be able to understand an easier one. Here is the approach: we first find the derivative of x + x + 4, which is x +. The idea is that we want to split the integral up into two pieces. We should be able to use the u-substitution u = x +x+4 for one of the pieces, and we should be able to use a trigonometric substitution like x = tan θ above for the other piece, so we want the other piece to have a constant numerator (meaning no x s in the numerator. We first divide the 6 in 6x + 7 by the in x + to get. Now we rewrite 6x + 7 as (x + + (the comes from subtracting 6x + 7 (x +. This tells us exactly how we will break up the integral: Now show that (6x + 7 (x + x + 4 = (x + (x + x = 4 + (x + x + 4 (x + (x + x + 4 = (x + x C. OK, all that s left is (x + x + 4, which is perhaps the trickiest. It doesn t look like an integral for which we could use trigonometric substitution, but it actually is. The thing is, we first have to complete the square: ( ( x + x + 4 = x + x

10 So = ( = ( + (x + x + 4 = ( + 4 (( + Now we can either use the substitution u = to make it look like an integral we solved in the last section, or we can just use the substitution = tan θ. Remember, if we have (x + c n/ in our integral, we need to use the substitution x = c tan θ; we have to take the square root of that constant c. And this constant ( in this case will always be positive precisely because the quadratic is irreducible, so we will always use a substitution involving tan θ. When you use the substitution = tan θ, you should get sec θ dθ ( sec θ When you solve this (by converting into cosines and using cos θ = +cos(θ and cos (θ = +cos(4θ you should get 8 θ + 4 sin(θ + sin(4θ + C Now we substitute back in. First, we solve for θ: ( θ = tan Second, we set up a triangle where x+ = tan θ. Namely, we put x+ on the opposite side of θ and on the side adjacent to θ. Then the hypotenuse is the familiar ( + ( = x + x + 4. Now, using trigonometric identities, and sin(θ = sin θ cos θ = x + x + 4 x + x + 4 0

11 sin(4θ = sin((θ = sin(θ cos(θ = sin(θ( cos θ ( = x + x + 4 x ( + x + 4 Thus the integral of this monster is (6x + 7 (x + x + 4 = 8 θ + 4 sin(θ + sin(4θ + C = ( 8 tan + ( 4 x + x + 4 x + x ( ( x + x + 4 x ( + x + 4 ( ( (x + x + 4 (x + x C Again, I emphasize that this is a much harder example than you will see on the test; it is even harder than any example on the homework, so don t worry!

Partial Fractions. (Do you see how to work it out? Substitute u = ax + b, so du = a dx.) For example, 1 dx = ln x 7 + C, x x (x 3)(x + 1) = a

Partial Fractions. (Do you see how to work it out? Substitute u = ax + b, so du = a dx.) For example, 1 dx = ln x 7 + C, x x (x 3)(x + 1) = a Partial Fractions 7-9-005 Partial fractions is the opposite of adding fractions over a common denominator. It applies to integrals of the form P(x) dx, wherep(x) and Q(x) are polynomials. Q(x) The idea