Integration of Rational Functions

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The cover-up method is a technique used to find the coefficients in the partial fraction decomposition of a rational function, $\frac{P(x)}{Q(x)}$. For this to work, the denominator $Q(x)$ must factor into distinct linear factors, and the degree of $P(x)$ must be strictly less than the degree of $Q(x)$.

Partial fraction decomposition: Distinct linear factors

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

Here is the general step-by-step process, using $\int \frac{4x-1}{x^2+x-2} dx$ as an example:

1. Factor the denominator of the rational function into linear and irreducible quadratic factors: $\int \frac{4x-1}{x^2+x-2}dx = \int \frac{4x-1}{(x-1)(x+2)}dx$
2. Write or express the rational function as a sum of partial fractions with unknown constants: $\frac{4x-1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}$.
3. Solve for the constants using the cover-up method:
   $\star$ Solve for A by multiplying the entire equation by (x-1), $\frac{4x-1}{(x+2)} = A + \frac{B(x-1)}{(x+2)}$.
   Set $x = 1$ (the root of the factor we multiplied by):
   $$\frac{4(1)-1}{1+2} = A + 0 \implies \frac{3}{3} = A \implies A = 1$$ $\star$ Solve for B: Multiply the entire equation by $(x+2)$: $\frac{4x-1}{x-1} = \frac{A(x+2)}{x-1} + B$ Set $x = -2$: $\frac{4(-2)-1}{-2-1} = 0 + B \implies \frac{-9}{-3} = B \implies B = 3$
4. Integrate each term obtained from the partial fraction decomposition, $\int \frac{4x-1}{x^2+x-2} dx = \int \frac{1}{x-1} dx + \int \frac{3}{x+2} dx = \ln\vert{}x-1\vert{} + 3\ln\vert{}x+2\vert{} + C$ where C is the constant of integration.

• Evaluate $\int \frac{1}{(x-2)(x-3)} dx$

1. Set up the decomposition: $\frac{1}{(x-2)(x-3)} = \frac{A}{x-2} + \frac{B}{x-3}$
2. Solve for A: Multiply by $(x-2)$ and set $x = 2$:
   $\frac{1}{2-3} = A \implies A = -1$
3. Solve for B: Multiply by $(x-3)$ and set $x = 3$:
   $\frac{1}{3-2} = B \implies B = 1$
4. Integrate: $\int \frac{1}{(x-2)(x-3)} dx = -\int \frac{1}{x-2} dx + \int \frac{1}{x-3} dx = -\ln\vert{}x-2\vert{} + \ln\vert{}x-3\vert{} + C$

Partial fraction decomposition: Repeated linear factors

When the denominator $Q(x)$ has repeated linear factors, the decomposition requires a term for every power of that factor.

• Evaluate $\int \frac{x^2+2}{x^3-3x+2} dx$

1. First, factor the denominator: $x^3-3x+2 = (x-1)^2(x+2)$.
   The partial fraction decomposition takes the form: $\frac{x^2+2}{(x-1)^2(x+2)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+2}$
2. Solve for C: Multiply by $(x+2)$ and set $x = -2$:
   $C = \frac{(-2)^2+2}{(-2-1)^2} = \frac{6}{9} = \frac{2}{3}$
3. Solve for B: Multiply by $(x-1)^2$ and set $x = 1$:
   $B = \frac{1^2+2}{1+2} = \frac{3}{3} = 1$
4. Solve for A: The cover-up method doesn't directly isolate A. Instead, substitute a convenient value for $x$ (like $x = 0$) and plug in the known values for B and C:
   $\frac{0^2+2}{(0-1)^2(0+2)} = \frac{A}{0-1} + \frac{1}{(0-1)^2} + \frac{2/3}{0+2}$
   $\frac{2}{2} = -A + 1 + \frac{1}{3} \implies 1 = -A + \frac{4}{3} \implies A = \frac{1}{3}$
5. Integrate: $\int \frac{x^2+2}{(x-1)^2(x+2)} dx = \int \left( \frac{1/3}{x-1} + \frac{1}{(x-1)^2} + \frac{2/3}{x+2} \right) dx = \frac{1}{3}\ln\vert{}x-1\vert{} - \frac{1}{x-1} + \frac{2}{3}\ln\vert{}x+2\vert{} + C$

• Evaluate $\int \frac{x+1}{(x-1)^3(x-2)} dx$

1. Set up the decomposition:
   $\frac{x+1}{(x-1)^3(x-2)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{(x-1)^3} + \frac{D}{x-2}$
2. Solve for C: Multiply by $(x-1)^3$ and set $x = 1$:
   $C = \frac{1+1}{1-2} = -2$
3. Solve for D: Multiply by $(x-2)$ and set $x = 2$:
   $D = \frac{2+1}{(2-1)^3} = 3$
4. To find A and B, multiply the entire original equation by the common denominator to clear fractions:
   $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 $
   Expand and group by powers of $x$:
   $x + 1 = (A + D)x^3 + (-4A + B - 3D)x^2 + (5A - 3B + C + 3D)x + (-2A + 2B - 2C - D)$
   By equating the coefficients on both sides, we get a system of equations. Let’s use the $x^3$ and $x^2$ coefficients:
   $x^3$ term: $A + D = 0 \implies A + 3 = 0 \implies A = -3$
   $x^2$ term: $-4A + B - 3D = 0 \implies -4(-3) + B - 3(3) = 0 \implies 12 + B - 9 = 0 \implies B = -3$
5. Integrate: $\int \left( \frac{-3}{x-1} - \frac{3}{(x-1)^2} - \frac{2}{(x-1)^3} + \frac{3}{x-2} \right) dx = -3\ln\vert{}x-1\vert{} + \frac{3}{x-1} + \frac{1}{(x-1)^2} + 3\ln\vert{}x-2\vert{} + C$

Partial Fraction Decomposition: Irreducible Quadratic Factors

When dealing with a quadratic factor that cannot be factored further over the real numbers (like $x^2+1$), the numerator for that fraction must be a linear term (e.g., $Bx + C$).

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

1. Factor the denominator by grouping: $x^2(x-1) + 1(x-1) = (x-1)(x^2+1)$.
   Set up the decomposition: $\frac{x^2}{(x-1)(x^2+1)} = \frac{A}{x-1} + \frac{Bx + C}{x^2+1}$
2. Solve for A: Multiply by $(x-1)$ and set $x = 1$:
   $A = \frac{1^2}{1^2+1} = \frac{1}{2}$
3. Solve for B and C:
   Clear the denominators (that is, multiply the entire original equation by the common denominator to clear fractions): $x^2 = A(x^2+1) + (Bx+C)(x-1)$
   Substitute $A = 1/2$ and expand: $x^2 = \frac{1}{2}x^2 + \frac{1}{2} + Bx^2 - Bx + Cx - C$
   Equate coefficients:
   $\star$ $x^2$ terms: $1 = \frac{1}{2} + B \implies B = \frac{1}{2}$
   $\star$ Constant terms: $0 = \frac{1}{2} - C \implies C = \frac{1}{2}$
4. Integrate: $\int \left( \frac{1/2}{x-1} + \frac{\frac{1}{2}x + \frac{1}{2}}{x^2+1} \right) dx = \frac{1}{2} \int \frac{1}{x-1} dx + \frac{1}{2} \int \frac{x}{x^2+1} dx + \frac{1}{2} \int \frac{1}{x^2+1} dx = \frac{1}{2}\ln\vert{}x-1\vert{} + \frac{1}{4}\ln(x^2+1) + \frac{1}{2}\arctan(x) + C$
   where $\int \frac{x}{x^2+1} dx = \frac{1}{2}\ln(x^2+1)$ using $u = x^2+1$

Integration of Improper Rational Fractions

• $\int \frac{x^2}{x+1}dx$

Polynomial Long Division:

          x - 1   (Quotient)
        -------
x + 1  |  x² + 0x + 0 (Divisor)
        -(x² + x)
       ---------
              - x + 0
            -(- x - 1)
            ---------
                    1   (Remainder)

Rewrite and integrate: $\int \frac{x^2}{x+1} dx = \int \left( x - 1 + \frac{1}{x+1} \right) dx = \frac{1}{2}x^2 - x + \ln\vert{}x+1\vert{} + C$

• Evaluate: $\int \frac{x}{x+1} dx$

You can use long division, or a quick algebraic trick (add and subtract 1 in the numerator):
$\int \frac{x+1-1}{x+1} dx = \int \left( 1 - \frac{1}{x+1} \right) dx = x - \ln\vert{}x+1\vert{} + C$

• Evaluate: $\int \frac{x^3}{x^2+x-2} dx$

             x - 1   (Quotient)
           -----------
x² + x - 2 | x³ + 0x² + 0x + 0 (Divisor)
             -(x³ + x² - 2x)
          ---------------
               - x² + 2x + 0
             -(- x² - x  + 2)
             ---------------
                  3x - 2   (Remainder)

1. Polynomial Long Division, then rewrite the integral: $\int \frac{x^3}{x^2+x-2} dx = \int \left( x - 1 + \frac{3x - 2}{x^2+x-2} \right) dx$
2. Now, apply partial fractions to the remainder term $\frac{3x - 2}{(x+2)(x-1)}$: $\frac{3x - 2}{(x+2)(x-1)} = \frac{A}{x+2} + \frac{B}{x-1}$.
3. Solve for A: Multiply by $(x+2)$, set $x = -2$: $A = \frac{3(-2)-2}{-2-1} = \frac{-8}{-3} = \frac{8}{3}$
4. Solve for B: Multiply by $(x-1)$, set $x = 1$: $B = \frac{3(1)-2}{1+2} = \frac{1}{3}$
5. Final Integration: $\int \left( x - 1 + \frac{8/3}{x+2} + \frac{1/3}{x-1} \right) dx = \frac{1}{2}x^2 - x + \frac{8}{3}\ln\vert{}x+2\vert{} + \frac{1}{3}\ln\vert{}x-1\vert{} + C$

Bibliography

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, Andrew Misseldine, blackpenredpen, and MathMajor, YouTube’s channels.
6. Contemporary Abstract Algebra, Joseph, A. Gallian.
7. MIT OpenCourseWare, 18.01 Single Variable Calculus, Fall 2007 and 18.02 Multivariable Calculus, Fall 2007, YouTube.
8. Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.