Integration of Algebraic Fractions 
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The method of partial fractions can be used in the integration of a proper algebraic fraction. This technique allows the integration to be done as a sum of much simpler integrals.
A proper algebraic fraction is a fraction of two polynomials whose top line is a polynomial of lower degree than the one on the bottom line. Recall that, for a polynomial x, the degree is the highest power of x. For example:
is a proper algebraic fraction because the top line is a polynomial of degree 1 and the bottom line is a polynomial of degree 2.
To integrate an improper algebraic fraction, one first needs to write the fraction as a sum of proper fractions. This first step can be done by using polynomial division (See section on Polynomial Division later in this module).
Look out for cases of proper algebraic fractions whose top line is a multiple k of the derivative of the bottom line. Then the standard integral:
can be used instead of working out partial fractions.
Otherwise, the bottom line of a proper algebraic fraction needs to be factorised as far as possible. This allows us to identify the form of each partial fraction needed:
factor in the bottom line  →  form of partial fraction(s) 

(ax + b)  Aax + b  
(ax + b)^{2}  Aax + b + B(ax + b)^{2}  
ax^{2} + bx + c  Ax + Bax^{2} + bx + c 
 where A and B are constants to be determined.
Perform the following integrations.
Click on the questions to reveal their solution
Exercise 1:
Solution:
The top line is quadratic in x.
The bottom line is linear in x.
Thus we have an improper algebraic fraction. We thus use simple (i.e. termbyterm) polynomial division.
∴ ∫ x^{2} + 2x + 5x dx  =  ∫ ( x^{2}x + 2xx + 5x ) dx 
=  ∫ ( x + 2 + 5x ) dx  
=  ∫ x dx + ∫ 2 dx + 5 ∫ 1x dx  
=  12 x^{2} + 2x + 5 ln x + C 
 where C is a constant of integration.
Exercise 2:
Solution:
The top line is cubic in x.
The bottom line is quadratic in x.
Thus we have an improper algebraic fraction. We thus use simple (i.e. termbyterm) polynomial division.
∴ ∫ x^{3} + 4x^{2} + 3x + 1x^{2} dx  =  ∫ ( x^{3}x^{2} + 4x^{2}x^{2} + 3xx^{2} + 1x^{2} ) dx 
=  ∫ ( x + 4 + 3x + 1x^{2} ) dx  
=  ∫ x dx + ∫ 4 dx + 3 ∫ 1x dx + ∫ x^{−2} dx  
=  12 x^{2} + 4x + 3ln x − 1x + C 
 where C is a constant of integration.
Exercise 3:
Solution:
The top line is quadratic in x.
The bottom line is linear in x.
Thus we have an improper algebraic fraction. We thus use full polynomial division. There is extra help with polynomial division in the Polynomial Division section. Here, we will go through the polynomial division first before the integration.
x^{2} + 3x + 4x + 1  =  x(x + 1)−x + 3x + 4x + 1 
=  x(x + 1) + 2x + 4x + 1  
=  x(x + 1)x + 1 + 2x + 4x + 1  
=  x + 2x + 4x + 1  
=  x + 2(x + 1)−2 + 4x + 1  
=  x + 2(x + 1) + 2x + 1  
=  x + 2(x + 1)x + 1 + 2x + 1  
=  x + 2 + 2x + 1 
Polynomial division is complete since we no longer have any improper algebraic fractions.
∴ ∫ x^{2} + 3x + 4x + 1 dx  =  ∫ ( x + 2 + 2x + 1 ) dx 
=  12 x^{2} + 2x + 2 ln x + 1 + C 
 where C is a constant of integration.
Exercise 4:
Solution:
The top line is quadratic in x.
The bottom line is linear in x.
Thus we have an improper algebraic fraction. We thus use full polynomial division.
2x^{2} + 5x + 2x + 2  =  2x(x + 2)−4x + 5x + 3x + 2 
=  2x(x + 2) + x + 3x + 2  
=  2x(x + 2)x + 2 + x + 3x + 2  
=  2x + x + 3x + 2  
=  2x + (x + 2)−2 + 3x + 2  
=  2x + (x + 2) + 1x + 2  
=  2x + (x + 2)x + 2 + 1x + 2  
=  x + 1 + 1x + 2 
Polynomial division is complete since we no longer have any improper algenbraic fractions.
∴ ∫ 2x^{2} + 5x + 3x + 2 dx  =  ∫ ( 2x + 1 + 1x + 2 ) dx 
=  x^{2} + x + ln x + 2 + C 
 where C is a constant of integration.
Exercise 5:
Solution:
The top line is degree 3 in x.
The bottom line is degree 4 in x.
Thus we have a proper algebraic fraction. So, we could factorise the bottom line for partial fractions. BUT FIRST, check if it is of the form ∫ k g'(x)g(x) , where k is a constant.
If g(x) = x^{4} + 2x + 3 (the bottom line), g'(x) = 4x^{3} + 2 (which exactly equals the top line). So we can use the standard integral
(or employ substitution techniques by setting u = x^{4} + 2x + 3)
Exercise 6:
Solution:
The top line is degree 1 in x.
The bottom line is degree 2 in x.
Thus we have a proper algebraic fraction. So, we could factorise the bottom line for partial fractions. BUT FIRST, check if it is of the form ∫ k g'(x)g(x) , where k is a constant.
If g(x) = x^{2}−5 (the bottom line), g'(x) = 2x (which is double the top line). So we can use the standard integral:
(or employ substitution techniques by setting u = x^{2} + − 5)
Exercise 7:
Solution:
The top line is degree 1 in x.
The bottom line is degree 2 in x.
Thus we have a proper algebraic fraction, and the top line is not a multiple of the derivative of bottom line. We thus try partial fractions.
17−x(x − 3)(x + 4)  =  A(x − 3) + B(x + 4) 
=  A(x + 4) + B(x − 3)(x − 3)(x + 4) 
∴ 17 − x = A(x + 4) + B(x − 3) [if true then true for all x]
x = −4 gives 17 + 4 = 0 + (−4 − 3)B, i.e. 21 = − 7B, B = −3
x = 3 gives 17 − 3 = (3 + 4)A + 0, i.e. 14 = 7A, A = 2
∴ ∫ 17−x(x − 3)(x + 4) dx  =  ∫ 2(x − 3) + (−3)(x + 4) dx 
=  2 ∫ dx(x − 3) − 3 ∫ dx(x + 4)  
=  2 ln x − 3 − 3 ln x + 4 + C 
Note: In the above we have used: ∫ dxax + b = 1a ln ax + b + D
Exercise 8:
Solution:
The top line is degree 1 in x.
The bottom line is degree 2 in x.
Thus we have a proper algebraic fraction, and the top line is not a multiple of the derivative of bottom line. We thus try partial fractions:
11x + 18(2x + 5)(x − 7)  =  A(2x + 5) + B(x − 7) 
=  A(x − 7) + B(2x + 5)(2x + 5)(x − 7) 
∴ 11x + 18 = A(x − 7) + B(2x + 5)
x = 7 gives 77 + 18 = (14 + 5)B, i.e. 95 = 19B, B = 5
x = −5 ⁄ 2 gives (−55 ⁄2) + 18 = (−5 ⁄ 2 −7)A, i.e. 19 ⁄ 2 = 19 ⁄ 2 A, A = 1
∴ ∫ 11x + 18(2x + 5)(x − 7) dx  =  ∫ 1(2x + 5) + 5(x − 7) dx 
=  ∫ dx(2x + 5) + 5 ∫ dx(x − 7)  
=  (1 ⁄ 2) ln 2x + 5 + 5 ln x − 7 + C 
Note: In the above we have used ∫ dxax + b = 1a ln ax + b + D
Exercise 9:
Solution:
The top line is degree 1 in x.
The bottom line is degree 3 in x.
Thus we have a proper algebraic fraction, and the top line is not a multiple of the derivative of bottom line. We thus try partial fractions.
7x + 1(x + 1)(x − 2)(x + 3)  =  A(x + 1) + B(x − 2) + c(x + 3) 
=  A(x − 2)(x + 3) + B(x + 1)(x + 3)+C(x + 1)(x − 2)(x + 1)(x − 2)(x + 3) 
∴ 7x + 1 = A(x − 2)(x + 3) + B(x + 1)(x + 3) + C(x + 1)(x − 2)
x = −1 gives −6 = A(−3)(2), i.e. −6=−6A, A = 1
x = 2 gives 15 = B(3)(5), i.e. 15 = 15B, B = 1
x = −3 gives −20 = C(−2)(−5), i.e. −20 = 10C, C = −2
∴ ∫ 7x + 1(x + 1)(x − 2)(x + 3) dx  =  ∫ 1(x + 1) + 1(x − 2) − 2 1(x + 3) dx 
=  ln x + 1 + ln x − 2 + ln x + 3 + C 
Exercise 10:
Solution:
The top line is degree 1 in x.
The bottom line is degree 2 in x.
Thus we have a proper algebraic fraction, and the top line is not a multiple of the derivative of bottom line. We thus try partial fractions.
2x + 9(x + 5)^{2}  =  A(x + 5) + B(x + 5)^{2} 
=  A(x + 5) + B(x + 5)^{2} 
∴ 2x + 9 = A(x + 5) + B
x = −5 gives −10 + 9 = B, i.e. B = −1
x = 0 gives 9 = 5A + B = 5A −1, i.e. 10 = 5A, A = 2
∴ ∫ 2x + 9(x + 5)^{2} dx  =  ∫ 2(x + 5) + (−1)(x + 5)^{2} dx 
=  2 ∫ dx(x + 5) − ∫ dx(x + 5)^{2}  
=  2 ln x + 5 − ∫ (x + 5)^{−2}dx + C  
=  2 ln x + 5 − (x + 5)^{−1}(−1) + C  
=  2 ln x + 5 + 1(x + 5) + C 
Note: In the last integral we have used : ∫ (ax + b)^{n} = (ax + b)^{n + 1}n + 1 + C, (n ≠ 1)
Exercise 11:
Solution:
The top line is degree 1 in x.
The bottom line is degree 2 in x.
Thus we have a proper algebraic fraction, and the top line is not a multiple of the derivative of bottom line. We thus try partial fractions.
13x − 4(3x − 2)(2x + 1)  =  A(3x − 2) + B(2x + 1) 
=  A(2x + 1) + B(3x − 2)(3x − 2)(2x + 1) 
∴ 13x − 4 = A(2x + 1) + B(3x − 2)
x = −1 ⁄ 2 gives −13 ⁄ 2 − 4 = B(−3 ⁄ 2 − 2), i.e. −21 ⁄ 2 = −7 ⁄ 2 B, B = 3
x = 2 ⁄ 3 gives (26 ⁄3)−(12 ⁄ 3) = A(4 ⁄ 3 + 3 ⁄ 3), i.e. 14 ⁄ 3 = 7 ⁄ 3 A, A = 2
∴ ∫ 13x − 4(3x − 2)(2x + 1) dx  =  ∫ 2(3x − 2) + 3(2x + 1) dx 
=  2 ∫ dx(3x − 2) + 3 ∫ dx(2x + 1)  
=  2(1 ⁄ 3) ln 3x − 2 + 3 (1 ⁄ 2) ln 2x + 1 + C  
=  (2 ⁄ 3) ln 3x − 2 + (3 ⁄ 2) ln 2x + 1 + C  
=  2 ln x + 5 + 1(x + 5) + C 
Note: In the above we have used: ∫ dxax + b = 1a ln ax + b + D
Exercise 12:
Solution:
The top line is degree 1 in x.
The bottom line is degree 3 in x.
Thus we have a proper algebraic fraction, and the top line is not a multiple of the derivative of bottom line. We thus try partial fractions.
27x(x − 2)^{2}(x + 1)  =  A(x − 2) + B(x − 2)^{2} + C(x + 1) 
∴ 27x = A(x − 2)(x + 1) + B(x + 1) + C(x − 2)^{2}
x = 2 gives 54 = 3B, i.e. B = 18
x = −1 gives −27 = C(−3)^{2}, i.e. C = −3
x = 0 gives 0 = A(−2) + 18 + (−3)(4), i.e. A = 3
∴ ∫ 27x(x − 2)^{2}(x + 1) dx  =  ∫ 3(x − 2) + 18(x − 2)^{2} − 3(x + 1) dx 
=  3 ln x − 2 + 18 ∫ (x − 2)^{−2} dx − 3 ln x + 1 + D  
=  3 ln x − 2 − 18(x − 2) − 3 ln x + 1 + D 
Exercise 13:
Solution:
The top line is degree 2 in x.
The bottom line is degree 3 in x.
Thus we have a proper algebraic fraction, and the top line is not a multiple of the derivative of bottom line. We thus try partial fractions. Note that x^{2} + x + 1 does not give real linear factors, so the partial fraction Bx + Cx^{2} + x + 1 is thus used. We then have
3x^{2}(x − 1)(x^{2} + x + 1)  =  A(x − 1) + Bx + Cx^{2} + x + 1 
3x^{2} = A(x^{2} + x + 1) + (Bx + C)(x − 1)
x = 1 gives 3 = 3A, i.e. A = 1
x = 0 gives 0 = A − C, i.e. C = A = 1
x = −1 gives 3 = A(1−1+1) + (−B + C)(−2), i.e. B = 2
∴ ∫ 3x^{2}(x − 1)(x^{2} + x + 1) dx  =  ∫ 1(x − 1) + 2x + 1x^{2} + x + 1 dx 
=  ln x − 1 + ln x^{2} + x + 1 + D 
Note that the second part of the integral is of the form:
You can use formal long division to simplify an improper algebraic fraction. In this module, we use another technique (sometimes called 'algebraic juggling').
In each step of the technique, we rewrite the top line in a way that the algebraic fraction can be broken into two separate fractions, where a simplifying cancellation is forced to appear in the first of these two fractions.
The technique involves rewriting the topline term with the highest power of x using the expression from the bottom line.
The detail of how the method works is best illustrated with the following worked example:
x^{3} + 3x^{2} − 2x − 1x + 1  =  x^{2}(x + 1) − x^{2} + 3x^{2}−2x − 1x + 1  (the bottom line has been used to write x^{3} as x^{2}(x + 1) − x^{2}) 
=  x^{2}(x + 1) + 2x^{2} − 2x − 1x + 1  
=  x^{2}(x + 1)x + 1 + + 2x^{2} − 2x − 1x + 1  
=  x^{2} + + 2x^{2} − 2x − 1x + 1  
=  x^{2} + 2x(x + 1) − 2x − 2x − 1x + 1  (writing 2x^{2} as 2x(x + 1) − 2x)  
=  x^{2} + 2x(x + 1) − 4x − 1x + 1  
=  x^{2} + 2x(x + 1)x + 1 + − 4x − 1x + 1  
=  x^{2} + 2x + − 4x − 1x + 1  
=  x^{2} + 2x + − 4(x + 1) + 4 −1x + 1  (writing −4x as −4(x + 1) + 4)  
=  x^{2} + 2x + −4(x + 1) + 3x + 1  
=  x^{2} + 2x + −4(x + 1)x + 1 + 3x + 1  
=  x^{2} + 2x − 4 + 3x + 1 
We have now written the original improper algebraic fraction as a sum of terms that do not involve any further improper fractions.
ƒ(x)  ∫ƒ(x) dx  ƒ(x)  ∫ƒ(x) dx  

x^{n}  x^{n+1} n+1 (n ≠ −1)  [g(x)]^{n} g'(x)  [g(x)]^{n+1} n+1 (n ≠ −1)  
1x ^{}  ln x  g'(x)g(x)  ln g(x)  
e^{x}  e^{x}  a^{x}  a^{x}ln a (a > 0)  
sin x  −cos x  sinh x  cosh x  
cos x  sin x  cosh x  sinh x  
tan x  − ln cosx  tanh x  ln cosh x  
cosec x  ln tan x2  cosech x  ln tanh x2  
sec x  ln sec x + tan x  sec x  2 tan^{−1}e^{x}  
sec^{2} x  tan x  sec^{2} x  tanh x  
cot x  ln sin x  cot x  ln sinh x  
sin^{2} x  x2 − sin 2x4  sinh^{2} x  sinh 2x4 − x2  
cos^{2} x  x2 + sin 2x4  cosh^{2} x  sinh 2x4 + x2  
1a^{2} + x^{2}  1a tan ^{−1} xa (a > 0)  √a^{2} + x^{2}  a^{2}2 [ sinh^{−1}( xa ) + x √a^{2}−x^{2} a^{2} ]  
1a^{2} − x^{2}  12a ln a + xa − x (0 < x < a)  √a^{2}−x^{2}  a^{2}2 [ sin^{−1}( xa ) + x √a^{2}−x^{2} a^{2} ]  
1x^{2} − a^{2}  12a ln x − ax + a (x > a > 0)  √x^{2} − a^{2}  a^{2}2 [−cosh^{−1}( xa ) + x √x^{2}−a^{2} a^{2} ]  
1 √ a^{2} + x^{2}  ln x + √a^{2} + x^{2} a (a > 0)  
1 √ a^{2} − x^{2}  sin^{−1} xa (−a < x < a)  1 √ x^{2} − a^{2}  ln x + √x^{2} − a^{2} a (x > a > 0) 