Calculus Lia Vas Partial Fractions rational function is a quotient of two polynomial functions The method of partial fractions is a general method for evaluating integrals of rational function The idea behind this method is to write a rational function as a sum of simpler rational functions called the partial fractions and then to integrate each term Case Let us consider first rational functions of the form p(x) where p and q are polynomials q(x) such that the degree of p is smaller than the degree of q In this case, to find the partial fractions: Factor the denominator into a product of powers of linear terms ax + b and quadratic terms ax + bx + c The quadratic equation ax + bx + c 0 should have no real solutions otherwise you would be able to factor ax + bx + c into a product of two linear terms For each power of a linear term of the form (ax+b) k, introduce k partial fractions with unknown coefficients,, k : ax + b + (ax + b) + + k (ax + b) k 3 For each power of a quadratic term of the form (ax + bx + c) k, introduce k partial fractions with unknown coefficients,, k and B, B, B k : x + B ax + bx + c + x + B (ax + bx + c) + + kx + B k (ax + bx + c) k 4 Determine the unknown coefficients by combining the partial fractions into a single fraction Be careful when finding the least common denominator: note that it should be equal to the initial denominator q(x) Then equate the coefficients of the numerator you obtain with the coefficients of the initial numerator p(x) This should give you a system in all the unknown coefficients Note that the number of equations should match the number of unknowns When you determine the unknown coefficients, you have found the partial fractions you need for the integration 5 Write the given rational function as a sum of partial fractions from the previous step and integrate each partial fraction Case If the rational function is of the form p(x) where p and q are polynomials such that the q(x) degree of p is greater or equal to the degree of q, then use the long division to divide the polynomial p by the polynomial q If s is the resulting quotient and r is the remainder, then you can write p q s + r q
The polynomial s can be easily integrated term-by-term The degree of r will be smaller than the degree of q so the rational function r falls in case category Thus, you can reduce case into case q Practice Problems a) Write out the form of the partial fractions of the given function (do not determine the constants) 4 x (x ) 3 (x + ) 5 (x )(x + ) (x )(x + ) 6 3 (x ) (x + ) (x ) (x + ) b) Evaluate the integrals 5 8 x dx x (x + ) dx 6 3 x + x + dx 9 x 3 + x x 9 x x + 3x 0 dx 3 + 5x x x dx 4 + 3x x 3 + x x 3 x + dx x + x 3 x + x dx 7 0 3x 4x + 5 (x )(x + ) dx x 4x (x ) (x + ) dx dx c) Find the area of the given region if the area is finite Solutions Part a) x 0, 0 y (x + )(x + 3) The denominator factors as (x )(x + ) Since you have two linear terms (both on power ), you need just two partial fractions, one for each term So, the function decomposes as + B x x+ The denominator factors as (x )(x + )(x + ) So, you have three linear terms (all on power ) and you need three partial fractions The function decomposes as + B + C x x+ x+ 3 The denominator factors as (x ) (x + ) So you have two linear terms, the first one is with power and the second with power So, for the first linear term, you need two fractions and B ) For the second linear term, you need another fraction C So, the function x (x ) x+ decomposes as + B + C x (x ) x+ 4 There are two linear terms, the first one is with power 3 and the second with power So, the function decomposes as + B + C + D + E x (x ) (x ) 3 x+ (x+) 5 There is a linear term and a quadratic term (note that x + cannot be factored further since x + 0 has no real solutions) You need a linear term in the denominator of the fraction with x + in denominator So, the function decomposes as + Bx+C x x +
6 The function decomposes as + B + Cx+D + Ex+F x (x ) x + (x +) Part b) By problem in part a), the function has the partial fractions decomposition as + B x The common denominator is the product (x )(x + ) so that + B (x+)+b(x ) x x+ x+ (x )(x+) x++bx B Note that this fraction should be equal to the initial function (note that the (x )(x+) denominators are equal) Thus x + + Bx B (x )(x + ) (x )(x + ) linear terms (with x) are equal + B free terms (with no x) are equal B 3 x + + Bx B Note that now we obtained a system of two equations with two unknowns (note that number of equations matches the number of unknowns) Using eliminations of variables, you can solve for one variable in one equation and substitute that in the other equation For example B using the first equation The second equation becomes + 3 Then B + 5 Finally, using partial fractions the integral becomes dx + 5 dx ln x + 5 ln x + + c x x+ x 3 dx / dx + 5/ dx (x )(x+) x x+ The denominator factors as (x + 5)(x ) The partial fractions decomposition is + B x+5 x Find and B similarly as in previous problem and obtain that and B So, the integral becomes dx + dx ln x + 5 ln x + c x+5 x 3 Note that the degree of the numerator is equal (not smaller!) than the degree of the denominator That means that you will have to divide the polynomials first x x x + (x x) x + Thus, the quotient is and the remainder is x + The function x + x+ is equal to + x x x x The denominator factors as x x x(x ) and so the fraction x+ x+ It has the x x x(x ) partial decomposition + B Find and B similarly as in previous problem and obtain that x x and B So, the integral becomes x + dx ( + x+ )dx ( + )dx x x x x x x x ln x + ln x + c 4 The denominator factors as x 3 +x x (x+) So, the partial fractions needed are x + B x + C x+ Note that least common denominator is x (x+), and not the product of all three denominators x 3 (x + ) Thus, x + B x + C x + x(x + ) + B(x + ) + Cx x (x + ) x + x + Bx + B + Cx x (x + ) 3
Equate this fraction with the initial function 5x +3x x 3 +x Equating the terms with x, you get that + C 5 So C 5 Equating the terms with x, you get that + B 3 Equating the terms with no x, you get B B Plugging that in the second equation gives you 3 4 Then obtain C from the first equation C 5 5 3 The integral becomes 5x +3x dx dx + dx + 3 dx ln x + + 3 ln + x + c x 3 +x x x x+ x 5 The denominator is a linear term on the third power So, the partial fraction decomposition is x + + B (x + ) + C (x + ) (x + ) + B(x + ) + C x + x + + Bx + B + C 3 (x + ) 3 (x + ) 3 Equate this fraction with the initial function x (x+) 3 Equating the terms with x, you get that x +0x+0 (x+) 3 Equating the terms with x, you get that + B 0 Since, B Equating the terms with no x, you get + B + C 0 C B So, the integral is x (x+) 3 dx x+ dx + (x+) dx + (x+) 3 dx ln x + + x+ (x+) + c 6 Since the degree of the numerator is larger than the degree of the denominator, you need to divide the polynomials first x x + x 3 (x 3 + x) x Thus, the quotient is x and the remainder is x The function x3 is equal to x x Note x + x + that the last fraction is already in the form of a partial fraction The integral of this fraction can be evaluated using substitution u x + This gives you x dx x du du x + u x u ln u ln(x +) So, the initial integral is x 3 dx (x x )dx x + x + x ln(x +)+c 7 Note that x + is a quadratic term that cannot be factored in two linear terms (since x + 0 has no real solutions) So, the partial fractions decomposition is + Bx+C x x + x + Bx + C x + (x + ) + (Bx + C)(x ) (x )(x + ) Equate this with 3x 4x+5 (x )(x +) Equating the terms with x, you get that + B 3 So B 3 x + + Bx Bx + Cx C (x )(x + ) Equating the terms with x, you get that B + C 4 So C 4 + B Equating the terms with no x, you get C 5 Hence + + 5 4 Thus B 3 and C 3 4
The integral becomes 3x 4x+5 dx dx + x 3 dx dx + x dx 3 dx (x )(x +) x x + x x + x + The first integral is ln x For the second, use substitution u x + to get ln(x + ) The third integral comes straight from the formula for tan x Thus, the given integral is ln x + ln(x + ) 3 tan x + c 8 x +x+ x 3 +x x +x+ + Bx+C Similarly as in previous problems, compute that, B x(x +) x x + x+ x and C When integrating the second term each term,, represent it as + x + x + x + Hence x+ dx + dx dx + x dx + dx The first and third integrals evaluate x x + x x + x + directly by the integration formulas Use the substitution u x + for the second integral to obtain the final answer ln x + ln( + x ) + tan x + c + B + C Similarly as in previous problems, compute x x (x ) that, B and C Integrating each term by using the formula for the integral of for the first two term and the power rule for the last term, produces ln x ln x + c x x x+ 9 x+ x+ x 3 x +x x(x x+) x(x ) x 0 4x (x ) (x +) + B + Cx+D Similarly as in previous problems, compute that, B x (x ) x +, C, and D Integrate each term as follows x dx+ x + (x ) dx+ x + dx x dx (x ) x dx x + dx+ x + dx You can use substitution u x + for the third integral and the rest of the integrals evaluate directly using the formulas to produce the final answer ln x + x ln( + x ) + tan x + c Part c) Set up the integral which evaluates the area first The condition x 0 indicates the bounds of the integration 0 x < The second condition indicates the lower and upper y-curves: y is the upper and y 0 is the lower curve Thus, the area can be found as (x+)(x+3) ( ) 0 (x + )(x + 3) 0 dx 0 (x + )(x + 3) dx Find the antiderivative first Use partial fractions + B Similarly as in previous (x+)(x+3) x+ x+3 problems, find that and B The integral becomes dx dx + dx (x+)(x+3) x+ x+3 ln x + ln x + 3 + c Thus, the area is (ln x + ln x + 3 ) 0 Since both x + and x + 3 are positive on [0, ), you can write this formula with parenthesis instead of the absolute value signs (keeping the absolute value is also correct) Thus (ln(x + ) ln(x + 3)) 0 lim (ln(x + ) ln(x + 3)) (ln() ln(3)) x Presently, the antiderivative evaluated at the upper bound is of the form To be able to use L Hopital s rule, write the function ln(x + ) ln(x + 3) as ln x+ x+3 Then lim ln x + x x + 3 Thus the area is 0 (ln() ln(3)) 0405 x + ln lim x x + 3 ln 5 ln 0