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Integration of rational functions.

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Partial fraction decomposition: Distinct linear factors

$\int (\frac{1}{x-4} + \frac{2}{x+3})dx = ln|x-4| + 2ln|x+3| + C.$

The cover-up method is a technique used for finding the coefficients in the partial fraction decomposition of a rational function, say $\frac{P(x)}{Q(x)}$. The denominator of the rational function should factor into distinct linear factors and deg(P) < deg(Q).

  1. Factorize the denominator of the rational function into linear and irreducible quadratic factors, e.g., $\int \frac{4x-1}{x^2+x-2}dx = \int \frac{4x-1}{(x-1)(x+2)}dx$
  2. Write or express the rational function as a sum of partial fractions, e.g., $\frac{4x-1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}$.
  3. Solve for the constants, e.g., solve for A by multiplying by (x-1), $\frac{4x-1}{(x+2)} = A + \frac{B(x-1)}{(x+2)}$ ⇒[For the factor x-1, set x = 1] $\frac{4-1}{1+2} = A + 0$ ⇒ A = 1; solve for B by multiplying by (x +2), $\frac{4x-1}{(x-1)} = \frac{A}{x-1}(x+2) + B$ ⇒ [For the factor x+2, set x = -2] $\frac{-8-1}{-2-1} = B$ ⇒ B = 3.
  4. Integrate each term obtained from the partial fraction decomposition, $\int \frac{4x-1}{x^2+x-2}dx = \int \frac{1}{x-1}dx + \int \frac{3}{x+2}dx = ln|x-1|+3ln|x+2| + C$ where C is the constant of integration.

$\frac{1}{(x-2)(x-3)} = \frac{A}{x-2} + \frac{B}{x-3}$ ⇒[solve for A by multiplying by (x-2)] $\frac{1}{(x-3)} = A + \frac{B(x-2)}{x-3}$ ⇒[set x = 2] $\frac{1}{-1} = A$ ⇒ A = -1. Analogously, [solve for A by multiplying by (x-3)] $\frac{1}{(x-2)} = \frac{A(x-3)}{x-2} + B$ ⇒[set x = 3] B = 1.

$\int \frac{1}{(x-2)(x-3)}dx = -\int \frac{1}{x-2}dx +\int \frac{1}{x-3}dx = -log|x-2| + log|x-3| + C.$

Partial fraction decomposition: Repeated linear factors

Let’s see a more complicated example when the denominator Q has a repeated linear factor, e.g., $\int \frac{x^2+2}{x^3-3x+2}dx = \int \frac{x^2+2}{(x-1)^2(x+2)}dx$. The partial fraction decomposition is of the form: $\frac{x^2+2}{(x-1)^2(x+2)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+2}$.

To obtain C, we multiply by (x+2), $\frac{x^2+2}{(x-1)^2} = \frac{A(x+2)}{x-1} + \frac{B(x+2)}{(x-1)^2} + C$, then set x = -2, $\frac{(-2)^2+2}{(-2-1)^2} = C, C = \frac{2}{3}$.

To obtain B, we multiply by (x-1)2, $\frac{x^2+2}{(x+2)} = A(x-1) + B + \frac{C(x-1)^2}{x+2}$, then set x = 1, $\frac{1^2+2}{1+2} = B$ ⇒ B = 1. This method does not work for the constant A.

Finally, we set x = 0, $\frac{2}{(-1)^2·2} = \frac{A}{-1} + \frac{B}{(-1)^2}+\frac{C}{2}$ =[B = 1, C = 2/3] $1 = \frac{A}{-1} + \frac{1}{(-1)^2}+\frac{2/3}{2}$ ⇒ 1 = -A +1 +1/3, A = 13.

$\int \frac{x^2+2}{x^3-3x+2}dx = \int \frac{x^2+2}{(x-1)^2(x+2)} = \int \frac{1/3}{x-1} +\frac{1}{(x-1)^2} + \frac{2/3}{x+2} dx = \frac{1}{3}ln|x-1| -\frac{1}{x-1} + \frac{2}{3}ln|x+2| + C.$

$\frac{x+1}{(x-1)^3(x-2)}dx = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{(x-1)^3} + \frac{D}{x-2}$ [1]

To obtain C, we multiply by (x-1)3, $\frac{x+1}{(x-2)} = A(x-1)^2 + B(x-1) + C + \frac{D(x-1)^3}{x-2}$, then set x = 1, C = $\frac{1+1}{1-2} = -2.$

To obtain D, we multiply by (x-2), $\frac{x+1}{(x-1)^3} = \frac{A(x-2)}{x-1} + \frac{B(x-2)}{(x-1)^2} + \frac{C(x-2)}{(x-1)^3} + D$, then set x = 2, D = $\frac{2+1}{(2-1)^3}=3$

Finally, combining the terms of [1],

$x +1 = A(x-1)^2(x-2) + B(x-1)(x-2) + C(x-2) +D(x-1)^3 = A(x^3-4x^2 + 5x − 2) + B(x^2 − 3x + 2) + C(x − 2) + D(x^3 − 3x^2 + 3x − 1) = (A + D)x^3 + (−4A + B − 3D)x^2 + (5A − 3B + C + 3D)x − 2A + 2B − 2C − D$,

and obtain the following equations,

A + D = 0 ⇒ A = -D = -3
−4A + B − 3D = 0 ⇒ 12 +B -9 = 0 ⇒ B = -3. \

$\int \frac{x+1}{(x-1)^3(x-2)}dx = \int \frac{-3}{x-1}dx + \int \frac{-3}{(x-1)^2}dx + \int \frac{-2}{(x-1)^3}dx + \int \frac{3}{x-2}dx = -3log|x-1|+\frac{3}{x-1}+\frac{1}{(x-1)^2}+3log|x-2|+C.$

The partial fraction decomposition is of the form: $\frac{x^2}{(x-1)(x^2+1)} = \frac{A}{x-1}+\frac{Bx + C}{x^2 +1}$.

Cover up for A is quite the same process, by multiplying for (x -1), $\frac{x^2}{(x^2+1)} = A +\frac{(Bx + C)(x-1)}{x^2 +1}$, then set x = 1, $\frac{1^2}{1^2+1} = A ⇒ A = \frac{1}{2}.$

Cover up for B and C is done by clearing the denominator, $x^2 = A(x^2+1) + (Bx+C)(x-1) = \frac{1}{2}(x^2+1) + (Bx+C)(x-1)$. x2 term, $1 = \frac{1}{2} + B ⇒ B = \frac{1}{2}$, and the constant term $0 = \frac{1}{2} -C ⇒ C = \frac{1}{2}$.

$\int \frac{x^2}{(x-1)(x^2+1)}dx = \int \frac{\frac{1}{2}}{(x-1)}dx + \int \frac{\frac{x}{2}}{(x^2+1)}dx + \int \frac{\frac{1}{2}}{(x^2+1)}dx= \frac{1}{2}ln|x-1| + \frac{1}{4}ln(x^2+1) + \frac{1}{2}tan^{-1}x + C~ because~ \frac{d}{dx}ln(x^2+1)=\frac{2x}{x^2+1}.$

$\int \frac{x}{(x^2+4)^3}dx$ =[u = x2+4, du = 2xdx] = $\int \frac{1}{2(u)^3}du = \frac{1}{2}\frac{u^{-2}}{-2}+C = \frac{-1}{4}u^{-2}+C = \frac{-1}{4(x^2+4)^2}+C$

Integration of improper rational fraction.

An improper rational fraction refers to a rational function where the degree of the numerator is greater than or equal to the degree of the denominator, e.g., $\frac{x^2}{x+1}, \frac{x^3}{(x-1)(x+2)}=\frac{x^3}{x^2+x-2}$.

Let’s apply long division,

x + 1 - x x x ² ² - 1 - - x x x + t h 1 1 e q u t o h t e i e r n e t m a i n d e r

To calculate $\int \frac{x^3}{x^2+x-2}dx$, let’s apply long division,

x ² + x - 2 - x x x ³ ³ - - - 1 x x x ² ² ² + + + t 2 2 x 3 h x x x e - 2 - q 2 u o t i t e h n e t r e m a i n d e r

$\int \frac{x^3}{x^2+x-2}dx = \int x - 1 + \frac{3x -2}{x^2 +x -2} dx$

Let’s factor the denominator x2+x-2 = (x +2)(x -1), $\frac{3x -2}{x^2 +x -2} = \frac{A}{x+2} + \frac{B}{x-1}$. To calculate A and B, we multiply for (x+2) and (x-1) respectively. $\frac{3x -2}{x -1} = A + \frac{B(x+2)}{x-1}$, set x = -2, $\frac{-6-2}{-2-1} = A ⇒ A = \frac{8}{3}$. Similarly, $\frac{3x -2}{x +2} = A\frac{x-1}{x+2} + B$, set x = 1, B = $\frac{3-2}{1+2} = \frac{1}{3}$

$\int \frac{x^3}{x^2+x-2}dx = \int x - 1 + \frac{3x -2}{x^2 +x -2} dx = \int x - 1 + \frac{\frac{8}{3}}{x+2} + \frac{\frac{1}{3}}{x-1} dx = \frac{1}{2}x^2 -x + \frac{8}{3}ln|x+2| +\frac{1}{3}ln|x-1| + C.$

Bibliography

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License and is based on MIT OpenCourseWare [18.01 Single Variable Calculus, Fall 2007].
  1. NPTEL-NOC IITM, Introduction to Galois Theory.
  2. Algebra, Second Edition, by Michael Artin.
  3. LibreTexts, Calculus. Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
  4. Field and Galois Theory, by Patrick Morandi. Springer.
  5. Michael Penn, and MathMajor.
  6. Contemporary Abstract Algebra, Joseph, A. Gallian.
  7. YouTube’s Andrew Misseldine: Calculus. College Algebra and Abstract Algebra.
  8. MIT OpenCourseWare 18.01 Single Variable Calculus, Fall 2007 and 18.02 Multivariable Calculus, Fall 2007.
  9. Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.
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