Integration of rational function by partial fractions

INTEGRATION OF RATIONAL FUNCTION BY PARTIAL FRACTION

If the denominator can be resolved in to rational factors of first or second Integration of rational functions by partial fractions.

In this section, we show how to integrate any rational functions (a ratio of Polynomials) by expressing it as a sum of simpler fractions, called partial fractions, that we already know how to integrate to see how the method of partial fraction works in general, let consider a rational functions f(x) = P(x)/Q(x) where P and Q are polynomial its possible to express f as a sum of similar 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) a_nxⁿ + a_n‒1xⁿ⁻¹ + … + a₁x + a₀ where a ≠ 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 P by Q (by long division) until a remember R(x) is obtained such that deg(R) < deg(Q),

The division formula statement is f(x) = P(x) / Q(x) = S(x) + [R(x)/Q(x)]

case (i) The denominator Q(x) is a product of district linear factors

This means that we can write Q(x) = (a₁x + b₁)(a₂x + b₂) ... (a_kx + b_k) where no factor is repeated (and no factor is a constant multiple of other)

In this case the partial fraction theorem states that there exits constant A₁, A₂, ... A_k such that

R(x) / Q(x) = A₁/[a₁x+b₁] + A₂/[a₂x+b₂] + …. + A_k/[a_kx+b_k]

The constant can be determined by multiplying both sides by Q(x) and by equating the coefficient of x^k, k = 1,2, ... and the constant terms

case (ii) Q(x) is a product of linear factors, some of which are repeated

Suppose the first linear factor (a₁x + b₁) is repeated r times, that is (a₁x + b₁)^r occurs in the factorization of Q(x). Then the partial fractions can be written as

R(x) / Q(x) = A₁/[a₁x+b₁] + A₂/[a₂x+b₂] + …. + A_k/[a_kx+b_k]

case (iii) Q(x) is a contains a repeated irreducible quadratic factors

If Q(x) has the factor (ax² + bx + c)^r, where b²‒4ac < 0, then the partial fraction contains the sum of the terms of the forms

occurs in the partial fraction decomposition of R(x)/Q(x). Each of the terms in the partial fraction can be integrated by using a substitution or by first completing the square if necessary. If degree of the numerator is less than the degree of the denominator, then the method of partial fraction is to be used.

Example 75. Evaluate ∫ dx / [x²‒a²]

Solution:

Let I = ∫ dx/√(x²‒a²) = ∫ dx / [(x+a)(x‒a)]

Since the degree of the numerator is less than the degree of the denominator, we don't need to divide.

Writing the function using partial fraction, we have

Example 76. Evaluate ∫ [x+5 / x²+x‒2] dx

Solution:

Let I = ∫ [x+5 / x²+x‒2] dx

Since the degree of the numerator is less than the degree of the denominator, we don't need to divide.

Writing the function using partial fraction, we have

Example 77. Evaluate 

Solution:

Let I = 

Since the degree of the numerator is less than the degree of the denominator, we don't need to divide.

Writing the function using partial fraction, we have

Example 78. Evaluate 

Solution:

Let I = 

Since the degree of the numerator is less than the degree of the denominator, we don't need to divide.

Writing the function using partial fraction, we have

2x³ + 3x² ‒ 2x = x(2x² + 3x‒2)

= x(2x‒1)(x+2)

Example 79. Find 

Solution:

Let 1 = 

Since, the denominator of the numerator is less than the denominator of the numerator, we use the term partial fraction.

Example 80. Find ∫ (x³+3) / (x‒1) dx

Solution:

Let I = ∫ (x³+3) / (x‒1) dx

Since the degree of the numerator is greater than the degree of the denominator, we first divide x³ + 3 by x‒1.

Example 81. Find ∫ [x³ / (x‒1)(x‒2)] dx

Solution:

Here, the degree of the numerator is greater than the degree of the denominator.

Example 82. 

Example 83. Evaluate ∫ (x⁴−2x²+4x+1) / (x³‒x²‒x+1) dx, by using partial fraction.

Solution:

Since the degree of the numerator is higher than the degree of the denominator, first applying the long division, we get

Example 84. Evaluate 

Solution:

Since the degree of the numerator is not less than the degree of the denominator, we first divide and obtain

Here the quadratic equation 4x² ‒ 4x + 3 is irreducible because its discriminant is b²‒4ac = ‒32 < 0. This mean it can't be factorized so we don't need to use the partial fraction technique.

Therefore, the integral is written as

Example 85. Evaluate ∫ (x²+x+1) / (x²‒x‒1) dx

Solution:

Since the degree of the numerator is equal to the degree of the denominator, we first divide and obtain

EXERCISE

12. Evaluate the following integrals