integration of rational functions

Integration of rational functions, which are quotients of polynomials, is a fundamental concept in calculus. The general form of a rational function is:

$$ \frac{P(x)}{Q(x)} $$

where ( P(x) ) and ( Q(x) ) are polynomials.

The strategy for integrating such functions typically involves a few steps:

(1) Polynomial Division: If the degree of ( P(x) ) is greater than or equal to the degree of ( Q(x) ), divide ( P(x) ) by ( Q(x) ). This step simplifies the function into a polynomial (which can be integrated directly) plus a rational function where the degree of the numerator is less than the degree of the denominator.

$f(x) = \frac {p(x)}{q(x)} = t(x) +\frac {p'(x)}{q'(x)} $

Let’s consider a rational function

$$ \frac{P(x)}{Q(x)} = \frac{x^3 – 2x^2 + 5x – 3}{x – 1} $$

The division of the polynomial $ x^3 – 2x^2 + 5x – 3 $ by ( x – 1 ) results in a quotient of $ x^2 – x + 4 $ and a remainder of ( 1 ).

So, the division can be expressed as:

$$ \frac{x^3 – 2x^2 + 5x – 3}{x – 1} = x^2 – x + 4 + \frac{1}{x – 1} $$

Here, $ x^2 – x + 4 $ is the polynomial part of the division, and $ \frac{1}{x – 1} $ is the fractional remainder.

(2)Factorization: Factorize the denominator ( Q(x) ) into its irreducible factors. This is crucial for the next step, partial fraction decomposition.

Lets consider a rational function

$$ \frac {1}{x^3 +6x^2+11x+6}$$

We can factorize the denominator as $x^3 +6x^2+11x+6=(x+1)(x+2)(x+3)$

Therefore

$$ \frac {1}{x^3 +6x^2+11x+6}= \frac {1}{(x+1)(x+2)(x+3)}$$

(3) Partial Fraction Decomposition: The goal here is to express the rational function as a sum of simpler fractions whose denominators are the factors of ( Q(x) ) and whose numerators are usually constants or linear polynomials. This step is based on the principle that any rational function can be expressed as a sum of simpler fractions.

A. $\frac {px +q}{(x-a)(x-b)}= \frac {A}{x-a} + \frac {B}{x-b} $
and $ \int \frac {px +q}{(x-a)(x-b)} dx =\int \left \{ \frac {A}{x-a} + \frac {B}{x-b} \right \} dx$

B. $\frac {px^2 +qx + r}{(x-a)(x-b)(x-c)} = \frac {A}{x-a} + \frac {B}{x-b} + \frac {C}{x-c}$
and $ \int \frac {px^2 +qx + r}{(x-a)(x-b)(x-c)} dx =\int \left \{ \frac {A}{x-a} + \frac {B}{x-b} + \frac {C}{x-c} \right \} dx$

C $\frac {px +q}{(x-a)^2} = \frac {A}{x-a} + \frac {B}{(x-a)^2}$
and $ \int \frac {px +q}{(x-a)^2} dx =\int \left \{ \frac {A}{x-a} + \frac {B}{(x-a)^2} \right \} dx$

D $\frac {px^2 +qx + r}{(x-a)^2(x-c)}=\frac {A}{x-a} + \frac {B}{(x-a)^2} + \frac {C}{x-c}$
and $ \int \frac {px^2 +qx + r}{(x-a)^2(x-c)} dx =\int \left \{ \frac {A}{x-a} + \frac {B}{(x-a)^2} + \frac {C}{x-c} \right \} dx$

E. $ \frac {px^2 +q+r}{(x-a)(x^2 + bx +c)}= \frac {A}{x-a} + \frac {Bx +C}{x^2 + bx +c} $
and $ \int \frac {px^2 +q+r}{(x-a)(x^2 + bx +c)} dx =\int \left \{ \frac {A}{x-a} + \frac {Bx +C}{x^2 + bx +c} \right \} dx$
where $x^2 + bx +c$ is a irreducible quadratic

(4) Integration of Each Term: Once the rational function is decomposed into simpler fractions, integrate each term separately. The integration techniques for these simpler fractions often involve basic antiderivatives, logarithms, or inverse trigonometric functions, depending on the form of the fraction.

Some formulas to remember based on that

I. $\int \frac {1}{x^2 – a^2} dx = \frac {1}{2a} ln |\frac {x-a}{x+a}| + C$
Proof
$\frac {1}{x^2 – a^2} =\frac {1}{2a}[ \frac {1}{x-a} – \frac {1}{x+a}]$
So
$\int \frac {1}{x^2 – a^2} dx $
$=\frac {1}{2a}[ \int \frac {1}{x-a} dx – \int \frac {1}{x+a}]$
$= \frac {1}{2a}[ln |x-a| – ln |x+a| + C$
$=\frac {1}{2a} ln |\frac {x-a}{x+a}| + C$

II. B. $\int \frac {1}{a^2 – x^2} dx = \frac {1}{2a} ln |\frac {a+x}{a-x}| + C$
Proof
$\frac {1}{a^2 – x^2} =\frac {1}{2a}[ \frac {1}{a-x} + \frac {1}{a+x}]$
So
$\int \frac {1}{a^2 – x^2} dx $
$=\frac {1}{2a}[ \int \frac {1}{a-x} dx + \int \frac {1}{x+a}]$
$= \frac {1}{2a}[-ln |a-x| + ln |a+x| + C$
$=\frac {1}{2a} ln |\frac {a+x}{a-x}| + C$

(5)Recombination: Finally, combine the integrals of the simpler fractions to obtain the integral of the original rational function.

It’s important to note that while the steps above are generally applicable, the specific method of partial fraction decomposition and integration can vary greatly depending on the nature of the rational function, particularly the form of its denominator. Some cases might require more advanced techniques like completing the square or trigonometric substitution

III. $\int \frac {1}{x^2 + a^2} dx = \frac {1}{a} \tan ^{-1} (\frac {x}{a}) + C$
Proof
Put $x =a tan \theta$ then $dx= a sec^2 \theta d\theta$
Therefore
$\int \frac {1}{x^2 + a^2} dx$
$=\int \frac {asec^2 \theta}{a^2 tan^2 \theta + a^2} d\theta$
$=\frac {1}{a} \int d\theta= \frac {1}{a} \theta + C = \frac {1}{a} \tan ^{-1} (\frac {x}{a}) + C$

(6) Completing the square technique

Some times, you may not able to factorize the denominator normally, then you might consider using the Completing the square technique

Example 1

$\int \frac {1}{ax^2 + bx + c} dx$

This integral can be converted to the form I,II, III given above and can be evaluated using the formula

Example 2

$\int \frac {px + q}{ax^2 + bx + c} dx$
Here we can decompose this integral using the below formula
$px+q = A \frac {d}{dx} (ax^2 + bx + c) + B$
Then integral is converted as
$=A \int \frac {\frac {d}{dx} (ax^2 + bx + c)} {ax^2 + bx + c} dx + B \int \frac {1}{ax^2 + bx + c} dx $

The first integral is of the form
$\int \frac {f'(x)}{f(x)} dx$ which can be easily evaluated
The second integral is of the form of example 1
$\int \frac {1}{ax^2 + bx + c} dx$ which can be easily evaluated

Solved Examples of integration of rational functions

Question 1

$$ \int \frac{1}{x^2 – 4} \, dx $$

Solution

1. Factorize the Denominator:
   $$ x^2 – 4 = (x + 2)(x – 2) $$
2. Partial Fraction Decomposition:
   $$ \frac{1}{x^2 – 4} = \frac{A}{x + 2} + \frac{B}{x – 2} $$
   Find ( A ) and ( B ) by setting up equations. $$ =\frac{-1}{4(x + 2)} + \frac{1}{4(x – 2)} $$
3. Integrate Each Term:
   $$ \int \frac{-1}{4(x + 2)} \, dx + \int \frac{1}{4(x – 2)} \, dx = \frac{\log(x – 2)}{4} – \frac{\log(x + 2)}{4} $$
   So, the integral of (\frac{1}{x^2 – 4}) is:

$$ \frac{\log(x – 2)}{4} – \frac{\log(x + 2)}{4} + C $$

Question 2

$$ \int \frac{x}{x^2 + 2x + 2} \, dx $$

Solution

Complete the Square for the Denominator:
$$ x^2 + 2x + 2 = (x+1)^2 + 1 $$

After completing the square, the denominator of the rational function (\frac{x}{x^2 + 2x + 2}) becomes ((x + 1)^2 + 1). Therefore, the integral becomes:

$$ \int \frac{x}{(x + 1)^2 + 1} \, dx $$

Now we can use the above technique in (6)

$ x = A \frac {d}{dx} ((x + 1)^2 + 1) + B$
Solving for A and B, we get
$ x = \frac {1}{2} \frac {d}{dx} ((x + 1)^2 + 1) -1$
Thus
$ \int \frac{x}{(x + 1)^2 + 1} \, dx = \int \frac {1}{2} \frac {\frac {d}{dx} ((x + 1)^2 + 1)}{(x + 1)^2 + 1} – \frac {1}{(x + 1)^2 + 1} \; dx$

Integrating this expression, we obtain:

$$ \frac{\log(x^2 + 2x + 2)}{2} – \arctan(x + 1) + C $$

where ( C ) is the constant of integration.