Comprehensive Study Guide on Partial Fraction Decomposition and Integration Techniques

Partial Fraction Decomposition Overview and Algorithm

• Definition and Purpose:

  • Partial fraction decomposition is an algebraic technique used to rewrite rational functions $f(x) = \frac{P(x)}{Q(x)}$ as a sum of simpler rational fractions that are easier to integrate.

  • Introductory example: Given the identity         $\frac{2x - 3}{(x - 1)(x - 2)} = \frac{1}{x - 1} + \frac{1}{x - 2}$         Integrating the left-hand side yields:         $\int \frac{2x - 3}{(x - 1)(x - 2)}\,dx = \int \frac{1}{x - 1}\,dx + \int \frac{1}{x - 2}\,dx = \ln(|x - 1|) + \ln(|x - 2|) + C$

• General Strategy for Integrating Rational Functions:

  • Let $P(x)$ and $Q(x)$ be polynomials, defining the rational function $f(x) = \frac{P(x)}{Q(x)}$.

  • Step 1: Check Rational Function Type (Proper vs. Improper):

    • If $f(x)$ is improper (meaning $\deg(P(x)) \ge \deg(Q(x))$), perform polynomial long division to decompose the fraction into:             $f(x) = S(x) + \frac{R(x)}{Q(x)}$             where $S(x)$ is the quotient polynomial and $R(x)$ is the remainder polynomial such that $\deg(R(x)) < \deg(Q(x))$.

    • If $f(x)$ is proper (meaning $\deg(P(x)) < \deg(Q(x))$), proceed directly to Step 2.

  • Step 2: Factor the Denominator $Q(x)$ Fully:

    • Factor $Q(x)$ completely into linear factors of the form $(ax + b)$ and irreducible quadratic factors of the form $(ax^2 + bx + c)$ where the discriminant $\Delta = b^2 - 4ac < 0$.

  • Step 3: Apply Partial Fraction Decomposition:

    • Set up the algebraic decomposition according to one or more of the four distinct factor cases.

Case 1: Denominator with Distinct Linear Factors

• Condition:

  • The denominator $Q(x)$ is a product of non-repeating, distinct linear factors.

• General Decomposition Form:

  • Suppose $Q(x) = (a_1 x + b_1)(a_2 x + b_2) \dots (a_k x + b_k)$.

  • The decomposition format is:         $\frac{P(x)}{Q(x)} = \frac{A_1}{a_1 x + b_1} + \frac{A_2}{a_2 x + b_2} + \dots + \frac{A_k}{a_k x + b_k}$         where $A_1, A_2, \dots, A_k$ are real constants to be solved.

• Worked Example 1: Integration with Distinct Linear Factors:

  • Evaluate the integral:         $\int \frac{6x^2 - 10x + 8}{x^3 - x^2 - 9x + 9}\,dx$

  • Factor the denominator by grouping:         $Q(x) = x^3 - x^2 - 9x + 9 = x^2(x - 1) - 9(x - 1) = (x^2 - 9)(x - 1) = (x - 3)(x + 3)(x - 1)$

  • Set up partial fraction decomposition:         $\frac{6x^2 - 10x + 8}{(x - 3)(x + 3)(x - 1)} = \frac{A}{x - 3} + \frac{B}{x + 3} + \frac{C}{x - 1}$

  • Multiply through by the common denominator:         $6x^2 - 10x + 8 = A(x + 3)(x - 1) + B(x - 3)(x - 1) + C(x - 3)(x + 3)$

  • Solve for constants by substitution:

    • Substitute $x = 3$:             $6(3)^2 - 10(3) + 8 = A(6)(2) \implies 32 = 12A \implies A = \frac{8}{3}$

    • Substitute $x = -3$:             $6(-3)^2 - 10(-3) + 8 = B(-6)(-4) \implies 92 = 24B \implies B = \frac{23}{6}$

    • Substitute $x = 1$:             $6(1)^2 - 10(1) + 8 = C(-2)(4) \implies 4 = -8C \implies C = -\frac{1}{2}$

  • Integrate each term:         $\int \left( \frac{8/3}{x - 3} + \frac{23/6}{x + 3} - \frac{1/2}{x - 1} \right)\,dx = \frac{8}{3}\ln(|x - 3|) + \frac{23}{6}\ln(|x + 3|) - \frac{1}{2}\ln(|x - 1|) + C$

• Worked Example 2: Trigonometric Integration via Substitution:

  • Evaluate the integral:         $\int \frac{6\cos(x)}{2\sin^2(x) - 9\sin(x) + 20}\,dx$

  • Substitute $u = \sin(x)$, leading to $du = \cos(x)\,dx$:         $\int \frac{6}{2u^2 - 9u + 20}\,du$

• Worked Example 3: Radical Expression Integration:

  • Evaluate the integral:         $\int \frac{x}{\sqrt{x + 9}}\,dx$

  • Substitute $u = \sqrt{x + 9}$, giving $u^2 = x + 9 \implies x = u^2 - 9$ and $dx = 2u\,du$:         $\int \frac{u^2 - 9}{u} (2u\,du) = 2 \int (u^2 - 9)\,du = 2 \left( \frac{u^3}{3} - 9u \right) + C = \frac{2}{3}(x + 9)^{3/2} - 18\sqrt{x + 9} + C$

Case 2: Denominator with Repeated Linear Factors

• Condition:

  • The denominator $Q(x)$ contains a linear factor $(a_1 x + b_1)$ raised to a power $k > 1$

• General Decomposition Form:

  • For each repeated linear factor $(a_1 x + b_1)^k$, include a sum of $k$ partial fractions with increasing powers in the denominator:         $\frac{A_1}{a_1 x + b_1} + \frac{A_2}{(a_1 x + b_1)^2} + \frac{A_3}{(a_1 x + b_1)^3} + \dots + \frac{A_k}{(a_1 x + b_1)^k}$         where $A_1, A_2, \dots, A_k$ are real constants.

• Worked Example 1: Form Setup:

  • Write the partial fraction decomposition form for:         $f(x) = \frac{2x^2 + 9x - 12}{(3x - 1)(x + 5)^2}$

  • Form:         $\frac{2x^2 + 9x - 12}{(3x - 1)(x + 5)^2} = \frac{A}{3x - 1} + \frac{B}{x + 5} + \frac{C}{(x + 5)^2}$

• Worked Example 2: Integration of Improper Function with Repeated Factors:

  • Evaluate the integral:         $\int \frac{4x^3 - 2x^2 + 4x + 1}{x^3 - x^2 - x + 1}\,dx$

  • Step 1: Perform long division since $\deg(P) = \deg(Q) = 3$:         $\frac{4x^3 - 2x^2 + 4x + 1}{x^3 - x^2 - x + 1} = 4 + \frac{2x^2 + 8x - 3}{x^3 - x^2 - x + 1}$

  • Step 2: Factor denominator:         $x^3 - x^2 - x + 1 = x^2(x - 1) - 1(x - 1) = (x^2 - 1)(x - 1) = (x + 1)(x - 1)^2$

  • Step 3: Decomposition setup:         $\frac{2x^2 + 8x - 3}{(x + 1)(x - 1)^2} = \frac{A}{x + 1} + \frac{B}{x - 1} + \frac{C}{(x - 1)^2}$

  • Step 4: Solve coefficients:         $2x^2 + 8x - 3 = A(x - 1)^2 + B(x + 1)(x - 1) + C(x + 1)$

    • Substitute $x = 1$: $2(1)^2 + 8(1) - 3 = C(2) \implies 7 = 2C \implies C = \frac{7}{2}$

    • Substitute $x = -1$: $2(-1)^2 + 8(-1) - 3 = A(-2)^2 \implies -9 = 4A \implies A = -\frac{9}{4}$

    • Substitute $x = 0$: $-3 = A(1) - B(1) + C(1) \implies -3 = -\frac{9}{4} - B + \frac{7}{2} \implies B = \frac{17}{4}$

  • Step 5: Integrate:         $\int \left( 4 - \frac{9/4}{x + 1} + \frac{17/4}{x - 1} + \frac{7/2}{(x - 1)^2} \right)\,dx = 4x - \frac{9}{4}\ln(|x + 1|) + \frac{17}{4}\ln(|x - 1|) - \frac{7}{2(x - 1)} + C$

Case 3: Denominator with an Irreducible Quadratic Factor

• Condition:

  • The denominator $Q(x)$ contains a quadratic factor $ax^2 + bx + c$ with discriminant $\Delta = b^2 - 4ac < 0$.

• General Decomposition Form:

  • For each non-repeated irreducible quadratic factor $ax^2 + bx + c$, include a term with a linear numerator:         $\frac{Ax + B}{ax^2 + bx + c}$         where $A$ and $B$ are constants to be determined.

• Worked Example 1: Form Setups:

  • Part (a): Write the form for:         $f(x) = \frac{2x^2 + 4x + 1}{(x - 1)(x^2 + 9)}$

    • Since $x^2 + 9$ is irreducible ($\Delta = -36 < 0$):             $\frac{2x^2 + 4x + 1}{(x - 1)(x^2 + 9)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + 9}$

  • Part (b): Write the form for:         $f(x) = \frac{x^3 + x^2 + 1}{(3x - 8)(x^2 - 3x + 2)}$

    • Factor quadratic denominator $x^2 - 3x + 2 = (x - 1)(x - 2)$ (reducible linear factors):             $\frac{x^3 + x^2 + 1}{(3x - 8)(x - 1)(x - 2)} = \frac{A}{3x - 8} + \frac{B}{x - 1} + \frac{C}{x - 2}$

• Worked Example 2: Integration Involving Irreducible Quadratic Term:

  • Evaluate the integral:         $\int \frac{24x^2}{(x - 1)(x^2 + 7)}\,dx$

  • Decomposition setup:         $\frac{24x^2}{(x - 1)(x^2 + 7)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + 7}$

  • Clear fractions:         $24x^2 = A(x^2 + 7) + (Bx + C)(x - 1)$

  • Solve constants:

    • Substitute $x = 1$:             $24(1)^2 = A(8) \implies 8A = 24 \implies A = 3$

    • Expand and group coefficients:             $24x^2 = 3(x^2 + 7) + Bx^2 - Bx + Cx - C = (3 + B)x^2 + (C - B)x + (21 - C)$

    • Match coefficients:

      • $x^2$ coefficient: $3 + B = 24 \implies B = 21$

      • Constant term: $21 - C = 0 \implies C = 21$

  • Integrate:         $\int \left( \frac{3}{x - 1} + \frac{21x + 21}{x^2 + 7} \right)\,dx = 3\int \frac{1}{x - 1}\,dx + 21\int \frac{x}{x^2 + 7}\,dx + 21\int \frac{1}{x^2 + 7}\,dx$         $= 3\ln(|x - 1|) + \frac{21}{2}\ln(x^2 + 7) + \frac{21}{\sqrt{7}}\arctan\left(\frac{x}{\sqrt{7}}\right) + C$

Case 4: Denominator with Repeated Irreducible Quadratic Factors

• Condition:

  • The denominator $Q(x)$ contains an irreducible quadratic factor $(ax^2 + bx + c)^k$ raised to a power $k > 1$, where $\Delta = b^2 - 4ac < 0$

• General Decomposition Form:

  • For each repeated irreducible quadratic factor $(ax^2 + bx + c)^k$, include a sum of $k$ linear terms with increasing powers in the denominator:         $\frac{A_1 x + B_1}{ax^2 + bx + c} + \frac{A_2 x + B_2}{(ax^2 + bx + c)^2} + \frac{A_3 x + B_3}{(ax^2 + bx + c)^3} + \dots + \frac{A_k x + B_k}{(ax^2 + bx + c)^k}$         where $A_1, \dots, A_k$ and $B_1, \dots, B_k$ are real constants.

• Worked Example 1: Form Setups:

  • Part (a): Write the form for:         $f(x) = \frac{x^4 + x^2 + 1}{(x^2 + x + 1)(x^2 + 4)^2}$

    • $x^2 + x + 1$ is irreducible ($\Delta = 1 - 4 = -3 < 0$).

    • $(x^2 + 4)^2$ is a repeated irreducible quadratic factor.

    • Form:             $\frac{x^4 + x^2 + 1}{(x^2 + x + 1)(x^2 + 4)^2} = \frac{Ax + B}{x^2 + x + 1} + \frac{Cx + D}{x^2 + 4} + \frac{Ex + F}{(x^2 + 4)^2}$

  • Part (b): Write the form for:         $f(x) = \frac{x^3 + x^2 + 1}{(x^4 + 10x^2 + 25)(x^2 - 3x + 2)}$

    • Factor $x^4 + 10x^2 + 25 = (x^2 + 5)^2$ (repeated irreducible quadratic).

    • Factor $x^2 - 3x + 2 = (x - 1)(x - 2)$ (distinct linear factors).

    • Form:             $\frac{x^3 + x^2 + 1}{(x^2 + 5)^2(x - 1)(x - 2)} = \frac{A}{x - 1} + \frac{B}{x - 2} + \frac{Cx + D}{x^2 + 5} + \frac{Ex + F}{(x^2 + 5)^2}$

• Worked Example 2: Integral Evaluation:

  • Evaluate the integral:         $\int \frac{1}{2x^2 - x - 3}\,dx$

  • Factor denominator: $2x^2 - x - 3 = (2x - 3)(x + 1)$

  • Decomposition setup:         $\frac{1}{(2x - 3)(x + 1)} = \frac{A}{2x - 3} + \frac{B}{x + 1}$

  • Clear fractions:         $1 = A(x + 1) + B(2x - 3)$

    • Substitute $x = -1$: $1 = B(-5) \implies B = -\frac{1}{5}$

    • Substitute $x = \frac{3}{2}$: $1 = A\left(\frac{5}{2}\right) \implies A = \frac{2}{5}$

  • Integrate:         $\int \left( \frac{2/5}{2x - 3} - \frac{1/5}{x + 1} \right)\,dx = \frac{1}{5}\ln(|2x - 3|) - \frac{1}{5}\ln(|x + 1|) + C = \frac{1}{5}\ln\left(\left|\frac{2x - 3}{x + 1}\right|\right) + C$