Using Linear Partial Fractions - AP Calculus BC
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What is the partial fraction form for a simple linear factor like $x-a$?
What is the partial fraction form for a simple linear factor like $x-a$?
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$\frac{A}{x-a}$. Each linear factor gets one partial fraction term.
$\frac{A}{x-a}$. Each linear factor gets one partial fraction term.
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What is the form of partial fractions for a repeated linear factor like $(x-a)^2$?
What is the form of partial fractions for a repeated linear factor like $(x-a)^2$?
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$\frac{A}{x-a} + \frac{B}{(x-a)^2}$. Repeated factors need terms for each power.
$\frac{A}{x-a} + \frac{B}{(x-a)^2}$. Repeated factors need terms for each power.
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Find $A$ in the decomposition $\frac{A}{x} + \frac{B}{x-1} = \frac{2x+3}{x(x-1)}$.
Find $A$ in the decomposition $\frac{A}{x} + \frac{B}{x-1} = \frac{2x+3}{x(x-1)}$.
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$A = 2$. Set $x=0$ to find $A$.
$A = 2$. Set $x=0$ to find $A$.
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What is a key benefit of partial fraction decomposition for solving integrals?
What is a key benefit of partial fraction decomposition for solving integrals?
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Breaks down complex rational functions. Makes integration much simpler.
Breaks down complex rational functions. Makes integration much simpler.
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What does each term in a partial fraction decomposition represent?
What does each term in a partial fraction decomposition represent?
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A simpler fraction with a linear denominator. Each term has constant numerator and linear denominator.
A simpler fraction with a linear denominator. Each term has constant numerator and linear denominator.
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What is the general form of a linear partial fraction decomposition?
What is the general form of a linear partial fraction decomposition?
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$\frac{A}{x-a} + \frac{B}{x-b}$. For distinct linear factors, each gets one term.
$\frac{A}{x-a} + \frac{B}{x-b}$. For distinct linear factors, each gets one term.
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Identify the partial fractions for $\frac{2x+1}{x^2+x-6}$.
Identify the partial fractions for $\frac{2x+1}{x^2+x-6}$.
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$\frac{A}{x-2} + \frac{B}{x+3}$. Factor $x^2+x-6 = (x-2)(x+3)$.
$\frac{A}{x-2} + \frac{B}{x+3}$. Factor $x^2+x-6 = (x-2)(x+3)$.
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Identify the partial fractions for $\frac{7x+9}{x^2+x}$.
Identify the partial fractions for $\frac{7x+9}{x^2+x}$.
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$\frac{A}{x} + \frac{B}{x+1}$. Factor $x^2+x = x(x+1)$.
$\frac{A}{x} + \frac{B}{x+1}$. Factor $x^2+x = x(x+1)$.
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Determine the partial fractions for $\frac{8x}{x^2-2x}$.
Determine the partial fractions for $\frac{8x}{x^2-2x}$.
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$\frac{A}{x} + \frac{B}{x-2}$. Factor $x^2-2x = x(x-2)$.
$\frac{A}{x} + \frac{B}{x-2}$. Factor $x^2-2x = x(x-2)$.
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Determine the partial fractions for $\frac{7x+4}{x^2-1}$.
Determine the partial fractions for $\frac{7x+4}{x^2-1}$.
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$\frac{A}{x-1} + \frac{B}{x+1}$. Factor $x^2-1 = (x-1)(x+1)$.
$\frac{A}{x-1} + \frac{B}{x+1}$. Factor $x^2-1 = (x-1)(x+1)$.
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What must be done if the rational function is improper for partial fractions?
What must be done if the rational function is improper for partial fractions?
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Perform polynomial long division first. Convert improper to proper fraction first.
Perform polynomial long division first. Convert improper to proper fraction first.
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Calculate $B$ in the decomposition $\frac{A}{x} + \frac{B}{x+1} = \frac{4}{x(x+1)}$.
Calculate $B$ in the decomposition $\frac{A}{x} + \frac{B}{x+1} = \frac{4}{x(x+1)}$.
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$B = -4$. Set $x=-1$ to isolate $B$.
$B = -4$. Set $x=-1$ to isolate $B$.
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Find the partial fraction decomposition of $\frac{2x+3}{(x-1)(x+2)}$.
Find the partial fraction decomposition of $\frac{2x+3}{(x-1)(x+2)}$.
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$\frac{A}{x-1} + \frac{B}{x+2}$. Factor $(x-1)(x+2)$, assign constants to each.
$\frac{A}{x-1} + \frac{B}{x+2}$. Factor $(x-1)(x+2)$, assign constants to each.
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Identify the partial fraction decomposition for $\frac{1}{x(x+1)}$.
Identify the partial fraction decomposition for $\frac{1}{x(x+1)}$.
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$\frac{A}{x} + \frac{B}{x+1}$. Factor denominator $x(x+1)$, then assign constants.
$\frac{A}{x} + \frac{B}{x+1}$. Factor denominator $x(x+1)$, then assign constants.
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Identify the partial decomposition for $\frac{3x+1}{(x-3)^2}$.
Identify the partial decomposition for $\frac{3x+1}{(x-3)^2}$.
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$\frac{A}{x-3} + \frac{B}{(x-3)^2}$. Repeated factor requires two terms.
$\frac{A}{x-3} + \frac{B}{(x-3)^2}$. Repeated factor requires two terms.
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What is the denominator structure required for partial fraction decomposition?
What is the denominator structure required for partial fraction decomposition?
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Product of linear or irreducible quadratic factors. Standard form required for partial fraction method.
Product of linear or irreducible quadratic factors. Standard form required for partial fraction method.
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What is the first step in decomposing a rational function into partial fractions?
What is the first step in decomposing a rational function into partial fractions?
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Factor the denominator. Essential for setting up partial fraction terms.
Factor the denominator. Essential for setting up partial fraction terms.
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What is the result of using partial fractions on $\frac{1}{(x+1)^2}$?
What is the result of using partial fractions on $\frac{1}{(x+1)^2}$?
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$\frac{A}{x+1} + \frac{B}{(x+1)^2}$. Repeated factor already factored, so direct decomposition.
$\frac{A}{x+1} + \frac{B}{(x+1)^2}$. Repeated factor already factored, so direct decomposition.
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Identify the partial fraction form for $\frac{1}{(x-1)(x-2)(x-3)}$.
Identify the partial fraction form for $\frac{1}{(x-1)(x-2)(x-3)}$.
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$\frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3}$. Three distinct linear factors need three terms.
$\frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3}$. Three distinct linear factors need three terms.
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What is the advantage of using partial fractions in integration?
What is the advantage of using partial fractions in integration?
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Simplifies the integral into manageable parts. Each simple fraction integrates easily.
Simplifies the integral into manageable parts. Each simple fraction integrates easily.
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What is the purpose of equating coefficients in partial fraction decomposition?
What is the purpose of equating coefficients in partial fraction decomposition?
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To solve for unknown constants. Matching coefficients determines constants.
To solve for unknown constants. Matching coefficients determines constants.
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Identify the partial fractions for $\frac{5x}{x^2+3x}$.
Identify the partial fractions for $\frac{5x}{x^2+3x}$.
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$\frac{A}{x} + \frac{B}{x+3}$. Factor out $x$ from denominator first.
$\frac{A}{x} + \frac{B}{x+3}$. Factor out $x$ from denominator first.
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State the partial fraction form for $\frac{6x+9}{x^3(x-1)}$.
State the partial fraction form for $\frac{6x+9}{x^3(x-1)}$.
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$\frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3} + \frac{D}{x-1}$. $x^3$ factor creates three terms plus linear factor.
$\frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3} + \frac{D}{x-1}$. $x^3$ factor creates three terms plus linear factor.
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What is the purpose of partial fraction decomposition in calculus?
What is the purpose of partial fraction decomposition in calculus?
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To simplify integration or differentiation. Converts complex fractions into simpler integrable forms.
To simplify integration or differentiation. Converts complex fractions into simpler integrable forms.
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Determine the partial fractions for $\frac{3x^2}{x^3-x}$.
Determine the partial fractions for $\frac{3x^2}{x^3-x}$.
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$\frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}$. Factor $x^3-x = x(x-1)(x+1)$.
$\frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}$. Factor $x^3-x = x(x-1)(x+1)$.
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Find the partial fractions for $\frac{x+6}{x^2-5x+6}$.
Find the partial fractions for $\frac{x+6}{x^2-5x+6}$.
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$\frac{A}{x-2} + \frac{B}{x-3}$. Factor $x^2-5x+6 = (x-2)(x-3)$.
$\frac{A}{x-2} + \frac{B}{x-3}$. Factor $x^2-5x+6 = (x-2)(x-3)$.
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Identify the partial fractions for $\frac{x^2+3x+2}{x^3-x}$.
Identify the partial fractions for $\frac{x^2+3x+2}{x^3-x}$.
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$\frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}$. Factor $x^3-x = x(x-1)(x+1)$.
$\frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}$. Factor $x^3-x = x(x-1)(x+1)$.
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Find $B$ in the decomposition $\frac{A}{x} + \frac{B}{x+2} = \frac{2x+5}{x(x+2)}$.
Find $B$ in the decomposition $\frac{A}{x} + \frac{B}{x+2} = \frac{2x+5}{x(x+2)}$.
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$B = 5$. Set $x=-2$ to isolate $B$.
$B = 5$. Set $x=-2$ to isolate $B$.
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Determine $A$ in the decomposition $\frac{A}{x} + \frac{B}{x+1} = \frac{3}{x(x+1)}$.
Determine $A$ in the decomposition $\frac{A}{x} + \frac{B}{x+1} = \frac{3}{x(x+1)}$.
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$A = 3$. Set $x=0$ to isolate $A$.
$A = 3$. Set $x=0$ to isolate $A$.
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What substitution is used to solve for $A$ in partial fractions?
What substitution is used to solve for $A$ in partial fractions?
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Set $x$ to make other term zero. Strategic substitution eliminates other terms.
Set $x$ to make other term zero. Strategic substitution eliminates other terms.
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