{"id":8926,"date":"2024-01-28T18:00:06","date_gmt":"2024-01-28T12:30:06","guid":{"rendered":"https:\/\/physicscatalyst.com\/article\/?p=8926"},"modified":"2024-01-28T18:00:11","modified_gmt":"2024-01-28T12:30:11","slug":"integration-of-rational-functions","status":"publish","type":"post","link":"https:\/\/physicscatalyst.com\/article\/integration-of-rational-functions\/","title":{"rendered":"integration of rational functions"},"content":{"rendered":"\n

Integration of rational functions, which are quotients of polynomials, is a fundamental concept in calculus. The general form of a rational function is:<\/p>\n\n\n\n

$$ \\frac{P(x)}{Q(x)} $$<\/p>\n\n\n\n

where ( P(x) ) and ( Q(x) ) are polynomials.<\/p>\n\n\n\n

The strategy for integrating such functions typically involves a few steps:<\/p>\n\n\n\n

(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.<\/p>\n\n\n\n

$f(x) = \\frac {p(x)}{q(x)} = t(x) +\\frac {p'(x)}{q'(x)} $<\/p>\n\n\n\n

Let’s consider a rational function<\/p>\n\n\n\n

$$ \\frac{P(x)}{Q(x)} = \\frac{x^3 – 2x^2 + 5x – 3}{x – 1} $$<\/p>\n\n\n\n

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 ).<\/p>\n\n\n\n

So, the division can be expressed as:<\/p>\n\n\n\n

$$ \\frac{x^3 – 2x^2 + 5x – 3}{x – 1} = x^2 – x + 4 + \\frac{1}{x – 1} $$<\/p>\n\n\n\n

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

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

Lets consider a rational function<\/p>\n\n\n\n

$$ \\frac {1}{x^3 +6x^2+11x+6}$$<\/p>\n\n\n\n

We can factorize the denominator as $x^3 +6x^2+11x+6=(x+1)(x+2)(x+3)$<\/p>\n\n\n\n

Therefore<\/p>\n\n\n\n

$$ \\frac {1}{x^3 +6x^2+11x+6}= \\frac {1}{(x+1)(x+2)(x+3)}$$<\/p>\n\n\n\n

(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.<\/p>\n\n\n\n

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$<\/p>\n\n\n\n

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$<\/p>\n\n\n\n

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$<\/p>\n\n\n\n

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$<\/p>\n\n\n\n

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<\/p>\n\n\n\n

(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.<\/p>\n\n\n\n

Some formulas to remember based on that<\/p>\n\n\n\n

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

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

(5)Recombination: Finally, combine the integrals of the simpler fractions to obtain the integral of the original rational function.<\/p>\n\n\n\n

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<\/p>\n\n\n\n

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

(6) Completing the square technique<\/p>\n\n\n\n

Some times, you may not able to factorize the denominator normally, then you might consider using the Completing the square technique<\/p>\n\n\n\n

Example<\/strong> 1<\/p>\n\n\n\n

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

This integral can be converted to the form I,II, III given above and can be evaluated using the formula<\/p>\n\n\n\n

Example 2<\/strong><\/p>\n\n\n\n

$\\int \\frac {px + q}{ax^2 + bx + c} dx$
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 $<\/p>\n\n\n\n

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<\/p>\n\n\n\n

Solved Examples of integration of rational functions<\/h2>\n\n\n\n

Question 1<\/strong><\/p>\n\n\n\n

$$ \\int \\frac{1}{x^2 – 4} \\, dx $$<\/p>\n\n\n\n

Solution<\/strong><\/p>\n\n\n\n

    \n
  1. Factorize the Denominator:
    $$ x^2 – 4 = (x + 2)(x – 2) $$<\/li>\n\n\n\n
  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. $$ =\\frac{-1}{4(x + 2)} + \\frac{1}{4(x – 2)} $$<\/li>\n\n\n\n
  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:<\/li>\n<\/ol>\n\n\n\n

    $$ \\frac{\\log(x – 2)}{4} – \\frac{\\log(x + 2)}{4} + C $$<\/p>\n\n\n\n

    Question 2<\/strong><\/p>\n\n\n\n

    $$ \\int \\frac{x}{x^2 + 2x + 2} \\, dx $$<\/p>\n\n\n\n

    Solution<\/strong><\/p>\n\n\n\n

    Complete the Square for the Denominator:
    $$ x^2 + 2x + 2 = (x+1)^2 + 1 $$<\/p>\n\n\n\n

    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:<\/p>\n\n\n\n

    $$ \\int \\frac{x}{(x + 1)^2 + 1} \\, dx $$<\/p>\n\n\n\n

    Now we can use the above technique in (6)<\/p>\n\n\n\n

    $ 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$<\/p>\n\n\n\n

    Integrating this expression, we obtain:<\/p>\n\n\n\n

    $$ \\frac{\\log(x^2 + 2x + 2)}{2} – \\arctan(x + 1) + C $$<\/p>\n\n\n\n

    where ( C ) is the constant of integration.<\/p>\n\n\n\n

    \n

    Other Integration Related Articles<\/p>\n\n\n\n

    Integration of modulus function<\/a><\/td>Integration of tan x<\/a><\/td><\/td><\/td><\/tr>
    Integration of sec x<\/a><\/td>Integration of greatest integer function<\/a><\/td><\/td><\/td><\/tr>
    Integration of trigonometric functions<\/a><\/td>Integration of cot x<\/a><\/td><\/td><\/td><\/tr>
    Integration of cosec x<\/a><\/td>Integration of sinx<\/a><\/td><\/td><\/td><\/tr>
    Integration of irrational functions<\/a><\/td>integration of root x<\/a><\/td><\/td><\/td><\/tr>
    Integration of fractional part of x<\/a><\/td>Integration of cos square x<\/a><\/td><\/td><\/td><\/tr><\/tbody><\/table><\/figure>\n<\/div><\/div>\n","protected":false},"excerpt":{"rendered":"

    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>\n","protected":false},"author":8,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_uag_custom_page_level_css":"","site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"set","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[498],"tags":[],"class_list":["post-8926","post","type-post","status-publish","format-standard","hentry","category-maths"],"yoast_head":"\nintegration of rational functions - physicscatalyst's Blog<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/physicscatalyst.com\/article\/integration-of-rational-functions\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"integration of rational functions - physicscatalyst's Blog\" \/>\n<meta property=\"og:description\" content=\"Integration of rational functions, which are quotients of polynomials, is a fundamental concept in calculus. 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