Integration By Partial Fractions

A function of the form P(x)/Q(x), where P(x) and Q(x) are polynomials, is called a rational algebraic function or a rational function of x. Consider the rational function (x+7)/(2x – 3)(3x + 4) = 1/2x–3 – 1/3x+4 

The two fractions on the RHS are called partial fractions. To integrate the rational function on the LHS, it is sufficient to integrate the two fractions on the RHS, which are easily integrable. This is known as the  method of partial fractions. In case the degree of P(x) (numerator) is not less than that of Q(x) (denominator), we carry out the division of P(x) by Q(x) and reduce the degree of the numerator.

In order to write P(x)/Q(x) in partial fractions, first of all we write 

Q(x) = (x – a)^k ... (x² + αx + β)^r ... where binomials are different, and then set

Note: Before proceeding to write a rational function as a sum of partial fractions, one must try to figure out whether it is a proper rational fraction or is rewritten as one.

A rational function P(x)/Q(x) is proper if the degree of polynomial Q(x) is greater than the degree of the polynomial P(x). In case the degree of P(x) greater than or equal to the degree of Q(x), we first write P(x)/Q(x) = h(x) + P(x)/Q(x), where h(x) is a polynomial and p(x) is a polynomial of degree less than the degree of polynomial Q(x). 

Similarly, the function P(x)/Q(x) is said to be an improper rational function if deg P(x) ≥ deg q(x).

We now discuss the various cases for solving questions by partial fractions:

Suppose we have a function of the form f(x)/g(x), then we have the following cases:

1. If the denominator is expressible as the product of non-repeating linear factors i.e.

Let g(x) = (x – a₁)(x – a₂)(x – a₃)........... (x – a_n).

In such a case we assume that

, where C₁, C₂, ...C_n are constants to be determined by equating the numerators of both the sides and then substituting the values of x as a₁, a₂, …, a_n.

2. If the denominator is expressible as the product of non-repeating linear factors i.e.

Let g(x) = (x-a)^k(x-a₁)(x-a₂)(x-a₃)........... (x-a_r).

In such a case we assume that

where C₁, C₂, ...C_k and D₁, D₂, ...D_r are constants to be determined.

Note: Some of these constants are determined by equating the numerators of both the sides as done in previous case. Remaining constants can be determined by equating the coefficients of like powers on both the sides.

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3. When some of the factors of denominator g(x) are quadratic but non-repeating.

Supose we have quadratic factors of the type ax² + bx + c, then the partial fractions should be assumed to be of the type (Ax + B)/(ax² + bx + c), where A and B are constants which can be computed by comparing the coeffcients of like powers of x in the numerators of both the sides. In fact, partial fractions in this case should better be assumed as

4. When some of the factors of the denominator g(x) are quadratic and repeating.

Corresponding to every quadratic repeating factor of the type (ax² + bx + c)^l, we assume 2l partial fractions of the type 

[A₀(2ax + b)/(ax² + bx + c) + A₁/(ax² + bx + c)] + [A₁(2ax + b)/(ax² + bx + c)² + A₂/(ax² + bx + c)²] + …................ + [A_{2l -1}(2ax + b)/(ax² + bx + c)^l + A_2l/(ax² + bx + c)^l].

Here, A, B and C are real numbers to be suitably determined.

We now discuss certain integrations using partial fractions:

Example 1:

Evaluate ∫ cos x dx/(1 + sin x)(2 + sin x).

Solution:      

Put sinx = t ⇒ cosx dx = dt

 = log (1 + t) – log (2 + t) + c

                                                         = log (1 + sin x)/(2 + sin x) + c

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Example 2:           

Evaluate ∫1/sin x (2 + cos x – 2 sin x) dx.

Solution:      

Put tan = t ⇒ sec² x/2 . 1/2 dx = dt

⇒ dx = 2/sec² x/2 dt

⇒ dx = 2/(1 + t²) dt, then we have

Expands into simple fractions

1 + t²/t(t–3) = A/t + B/(t–3) + C/(t–1)

After solving the coefficients, A = 1/3; B = 5/3; C = –1

             = 1/3 In |t| + 5/3 |t–3| – In |t–1| + c

Example 3: 

Evaluate ∫ x dx/(x³ + 1).

Solution: 

Since x³ + 1 = (x + 1) (x² - x + 1) (the second factor is not a product of linear factors), hence the partial fractions of the given integer will have the form.

Hence, x = A(x² - x + 1) + (Bx + D)(x + 1) = (A + B)x² + (-A + B + D)x + (A + D).

Thus l = – 1/3 ∫dx/(x +1) + 1/3 ∫(x+1)/(x² - x + 1) dx = – 1/3 |x + 1| + 1/3 l₁.   

To calculate the integral l₁ = ∫(x + 1)/(x² – x + 1)dx.

Express (x +1) = l (d.c. of x² - x + 1) + m 

⇒ (x + 1) = l(2x - 1) + m = 2xl - l + m

Comparing the coefficients of like powers of x, we get l =1/2 and m = 3/2

⇒ l₁ = 1/2 ∫(2x–1)/(x² - x + 1) dx + 3/2 ∫dx/(x² - x + 1).

Q1. The suitable partial fraction for the rational function of the form (px + q)/(x-a)(x-b) is

(a) C/(x-a) + D/(x-b)

(b) C/(x-a)(x-b)

(c) C/(x-a)² + D/(x-b)²

(d) none of these

Q2. In an improper rational function,

(a) degree of numerator = degree of denominator

(b) degree of numerator ≥ degree of denominator

(c) degree of numerator ≤ degree of denominator

(d) degree of numerator > degree of denominator

Q3. When the denominator is a combination of linear and quadratic factors, the suitable partial fractions are

Q3. When some of the factors of the denominator are quadratic and repeating such as (px² + qx + r)^t, then

(a) we assume ‘t’ partial fractions

(b) we assume ‘t/2’ partial fractions

(c we may assume ‘2t’ or ‘t’ partial fractions

(d) we assume ‘2t’ partial fractions

Q4. The partial fractions for the rational function (px² + qx + r)/(x-c)(x² + dx + e) are

(a)  A/(x - c) + (Bx² + C)/(x² + dx + e)

(b)  A/(x - c)² + (Bx + C)/(x² + dx + e)

(c) A/(x - c) + (Bx + C)/(x² + dx + e)

(d) None of these

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