We know that a Rational Number can be expressed in the form of p/q, where p and q are integers and q≠0. Similarly, a rational function is defined as the ratio of two polynomials which can be expressed in the form of partial fractions: P(x)/Q(x), where Q(x)≠0.
In this section, we show how to integer any rational function (a ratio of polynomials) by expressing it as a sum of simpler fractions, called partial fractions, that we already know how to integrate. To illustrate the method, observe that by taking the fractions 2/(x – 1) and 1/(x + 2) to a common denominator we obtain
If we now reverse the procedure, we see how to integrate the function on the right side of this equation
To see how the method of partial fractions works in general, let's consider a rational function
Where P and Q are polynomials. It's possible to express f as sum of simpler fractions provided that the degree of P is less than the degree of Q. Such a rational function is called proper. Recall that if P(x) =
where an ≠ 0, then the degree of P is n and we write deg (P) = n.
If f is improper, that is, deg(P) deg (Q), then we must take the preliminary step of dividing Q into P (by division) until a remainder R(x) is obtained such that deg (R) < deg(Q). The division statement is
where S and R are also polynomials. As the following example illustrates, sometimes this preliminary step is all that is required.
Ex.1 Evaluate dx.
Sol. Since the degree of the numerator is greater than the degree of the denominator, we first perform the long division. This enables us to write
The next step is to factor the denominator Q(x) as far as possible . It can be shown that any polynomial Q can be factored as a product of linear factors (of the form ax + b) and irreducible quadratic factors
(of the form ax^{2} + bx + c, where b^{2} – 4ac < 0).
For instance, if Q(x) = x^{4} – 16, we could factor it as Q(x) = (x^{2} – 4) (x^{2} + 4) = (x – 2) (x + 2) (x^{2} + 4)
The third step is to express the proper rational function R(x)/Q(x) (from equation 1) as a sum of partial fractions of the form
A theorem in algebra guarantees that it is always possible to do this. We explain the details for the four cases that occur.
This means that we can write Q(x) = (a_{1}x + b1) (a_{2}x + b2) ... (a_{k}x + bk)
where no factor is repeated (and no factor is a constant multiple of another). In this case the partial fraction theorem states that there exist constants A_{1},A_{2},...,A_{k} such that.
These constants can be determined as in the following example.
Ex.2 Evaluate dx.
Sol. Since the degree of the numerator is less than the degree of the denominator, we don't need to divide.
We factor the denominator as 2x^{3} + 3x^{2} – 2x = x(2x^{2} + 3x – 2) = x(2x – 1) (x + 2)
Since the denominator has three distinct linear factors, the partial fraction decomposition of the integrand (2) has the form.
To denominator the values of A, B and C, we multiply both sides of this equation by the product of the denominators, x(2x – 1) (x + 2), obtaining.
(4) x^{2} + 2x – 1 = A(2x – 1) (x + 2) + Bx(x + 2) + Cx(2x – 1)
Expanding the right side of equation 4 and writing it in the standard form for polynomials, we get
(5) x^{2} + 2x – 1 = (2A + B + 2C) x^{2} + (3A + 2B – C) x – 2A
The polynomials in Equation 5 are identical, so their coefficients must be equal. The coefficient of x^{2 }on the right side, 2A + B + 2C, must equal the coefficient of x^{2} on the left sidenamely, 1. Likewise. The coefficients of x are equal and the constant terms are equal. This gives the following system of equation for A, B and C.
2A + B + 2C = 1 , 3A + 2B – C = 2 , – 2A = –1
Solving, we get
In integrating the middle term we have made the mental substitution u = 2x – 1, which gives du = 2dx and dx = du/2.
(6)
By way of illustration, we could write
but we prefer to work out in detail a simpler example.
Ex.3 Evaluate dx
Sol.
The first step is to divide. The result of long division is
The second step is to factor the denominator Q(x) = x^{3} – x^{2} – x + 1. Since Q(1) = 0, we know that x – 1 is a factor and we obtain x^{3} – x^{2} – x + 1 = (x – 1) (x^{2 }– 1) = (x – 1) (x – 1)(x + 1) = (x – 1)^{2} (x + 1) Since the linear factor x – 1 occurs twice, the partial fraction decompositoin is
Multiplying by the least common denominator (x – 1)^{2} (x + 1), we get
(7) 4x = A(x – 1)(x + 1) + B(x + 1) + C(x – 1)^{2} = (A + C) x^{2} + (B – 2x) x + (–A + B + C)
Now we equate coefficients : A + C = 0,B – 2C = 4, –A + B + C = 0
Solving, we obtain A = 1, B = 2, and C = – 1, so
(8) where A and B are constants to be determined. For instance, the function givenby f(x) = x/[(x – 2)(x^{2} + 1) (x^{2} + 4)] has a partial fraction decomposition of the form
The term given in (8) can be integrated by completing the square and using the formula.
Ex.4 Evaluate dx
Sol.
Since x^{3} + 4x = x(x^{2} + 4) can't be factored further, we write
Multiplying by x(x^{2} + 4), we have 2x^{2} – x + 4 = A (x^{2} + 4) + (Bx + C) x = (A + B) x^{2} + Cx + 4A Equating coefficients, we obtain A + B =2 C = –1 4A = 4
Thus A = 1, B = 1 and C = –1 and
In order to integrate the second term we split it into to parts
We make the substitution u = x^{2} + 4 in the first of these integrals so that du = 2x dx. We evaluate the second integral by means of Formula 9 with a = 2.
Ex.5 Write out the form of the partial fraction decomposition of the function
Sol.
Ex.6 Evaluate
Sol. The form of the partial fraction decomposition is
Multiplying by x(x^{2} + 1)^{2}, we have –x^{3} + 2x^{2 }– x + 1 = A(x^{2} + 1)^{2} + (Bx + C) x (x^{2} + 1) + (Dx + E)x
= A(x^{4 }+ 2x^{2} + 1) + B(x^{4} + x^{2}) + C(x^{3} + x) + Dx^{2} + Ex = (A + B) x^{4} + Cx^{3} + (2A + B + D) x^{2} + (C + E) x + A
If we equate coefficient, we get the system A + B = 0 ,C = –1 ,2A + B + D = 2 ,C+E =–1 , A=1
Which the solution A = 1, B = –1, D = 1, and E = 0. Thus
We note that sometimes partial fractions can be avoided when integrating a rational function. For instance, although the integral
could be evaluated by the method of case III, it's much easier to observe that if u = x(x^{2 }+ 3) = x^{3} + 3x, then du = (3x^{2 }+ 3) dx and so
Ex.7 Evaluatedx
Sol. In this example there is a repeated quadratic polynomial in the denominator. Hence, according to our previous discussion
For some constants A_{1}, B_{1}, A_{2} and B_{2}
An easy way to determine these constant is as follows. By long division,
Thus A_{1 }= 1, B_{1} = – 3, A_{2} = 1 and B_{2 }= 0
Ex.8 Evaluate
Sol.
put sin x – cos x = s and sin x + cosx = t ⇒ (cosx + sinx) dx = ds and (cosx – sin x) dx = dt
Ex.9 Evaluate
Sol.
We have , [multiplying the Nr. and Dr. by e^{x}]
putting e^{x }= t so that e^{x} dx = dt.
1 ≡ A (t1)2 + Bt(t1)+Ct ....(1) (on resolving into partial fractions)
To find A, putting t = 0 on both sides of (1), we get A = 1
To find C, put t = 1 and we get C = 1. Thus 1 ≡ (t  1)^{2} + Bt (t  1) + t
Comparing the coefficients of t^{2} on both sides, we get 0 = 1 + B or B = – 1
= log t – log (t – 1) – {1/(t – 1)} + C
= log e^{x} – log (e^{x} – 1) – {1/(e^{x} – 1)} + c = x – log (e^{x} – 1) – {1/(e^{x} – 1)} + c
Ex.10 Integrate (3x + 1) / {(x – 1)^{3} (x + 1)}.
Sol.
Putting x – 1 = y so that x = 1 + y, we get
arranging the N^{r}. and the D^{r}. in ascending powers of y
Hence the required integral of the given fraction
Ex.11 Evaluate the integral
Sol.
Let x = sec^{2}θ ⇒ dx = 2 sec^{2}θ tanq dθ
Ex.12 Integrate,
Sol.
Ex.65 Integrate
Sol.
Put y = tanθ
Ex.13 Evaluate dx, where f(x) is a polynomials of degree 2 in x such that f(0) = f(1) = 3f(2) = –3.
Sol.
Let , f(x) = ax^{2} + bx + c given, f(0) = f(1) = 3f(2) = –3
∴ f(0) = f(1) = 3f(2) = –3, f(0) = c = –3, f(1) = a + b + c = – 3, 3f(2) = 3 (4a + b + c) = – 3
on solving we get a = 1, b = – 1, c = –3
∴ f(x) = x^{2} – x – 3
Using partical fractions, we get,
We get, A = –1, B = 2, C = 2
Ex.14 Integrate 1/(sinx + sin 2x).
Sol.
Now putting cos x = t, so that – sin x dx = dt, we get
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1. What is integration by partial fractions? 
2. When should I use integration by partial fractions? 
3. How do I decompose a rational function into partial fractions? 
4. Can integration by partial fractions be used for improper rational functions? 
5. Are there any special cases or restrictions when using integration by partial fractions? 

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