AP Calculus BC Flashcards: Integrating Long Division Completing The Square

What this deck covers

This deck focuses on Integrating Long Division Completing The Square, giving you a quick way to review the definitions, rules, and examples that matter most for AP Calculus BC.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Flashcard 1: Integrate $\frac{1}{x^2 - 4}$. What method simplifies it?

Answer: Partial fraction decomposition. Factor as $(x-2)(x+2)$ for partial fraction decomposition.

Flashcard 2: What result is obtained by integrating $\frac{1}{a^2 + x^2}$?

Answer: $\frac{1}{a} \arctan(\frac{x}{a}) + C$. Factor out $a^2$ to get $\frac{1}{a^2}\cdot\frac{1}{1+(x/a)^2}$ form.

Flashcard 3: What is the first step in integrating using long division?

Answer: Divide the numerator by the denominator. This creates polynomial and remainder terms for separate integration.

Flashcard 4: Complete the square: $4x^2 - 12x + 9$. What is the result?

Answer: $(2x-3)^2$. Recognize $4x^2 - 12x + 9$ as a perfect square trinomial.

Flashcard 5: Perform polynomial division: $\frac{x^3 - 6x^2 + 11x - 6}{x - 2}$.

Answer: $x^2 - 4x + 3$. Synthetic or long division of cubic by linear factor.

Flashcard 6: What is the purpose of completing the square in integration?

Answer: To transform a quadratic expression for easier integration. Creates standard forms like $(x-h)^2 + k$ for known integral formulas.

Flashcard 7: Find and correct the error: $x^2 + 6x + 8 = (x+3)^2 + 1$

Answer: Correct: $x^2 + 6x + 8 = (x+3)^2 - 1$. The constant term is $8 = 9 - 1$, not $9 + 1$.

Flashcard 8: Complete the square for $x^2 + 2x + 1$. What is the result?

Answer: $(x+1)^2$. Already a perfect square, no completing needed.

Flashcard 9: Perform long division: $\frac{x^4 + x^3 + x + 1}{x^2 + 1}$.

Answer: $x^2 + 1 + \frac{x}{x^2+1}$. Divide quartic by quadratic, getting quotient plus remainder fraction.

Flashcard 10: How do you integrate $\frac{x^2 - 2x + 3}{x - 1}$ using long division?

Answer: Divide, then integrate the result. Long division separates into polynomial plus simple fraction terms.

Flashcard 11: What integral results from $\frac{1}{(x-h)^2 + k}$?

Answer: $\frac{1}{\text{sqrt}(k)} \text{arctan} \frac{x-h}{\text{sqrt}(k)} + C$. General arctangent integral formula after completing the square.

Flashcard 12: Which integral results from $\frac{1}{x^2+1}$?

Answer: $\text{arctan}(x) + C$. Standard arctangent integral form.

Flashcard 13: Perform polynomial division on $x^3 + 3x^2 + 3x + 1$ by $x + 1$.

Answer: $x^2 + 2x + 1 + \frac{0}{x+1}$. Recognize this as $(x+1)^3$ expanded, divides evenly.

Flashcard 14: Why is completing the square helpful in integration?

Answer: Transforms to a form suitable for standard integrals. Creates forms matching known integral formulas like $\frac{1}{u^2+a^2}$.

Flashcard 15: What is the integral of $\frac{1}{x^2 - 9}$?

Answer: $\frac{1}{6} \ln\left| \frac{x-3}{x+3} \right| + C$. Partial fraction decomposition of $\frac{1}{(x-3)(x+3)}$.

Flashcard 16: Find the integral: $\frac{1}{(x+3)^2 + 9}$.

Answer: $\frac{1}{3} \arctan \frac{x+3}{3} + C$. Use $u = \frac{x+3}{3}$ substitution with $\frac{1}{u^2+1}$ form.

Flashcard 17: Perform long division: $\frac{x^3 + 2x^2 + x + 1}{x + 1}$.

Answer: $x^2 + x + 1 + \frac{0}{x+1}$. Divide $x^3 + 2x^2 + x + 1$ by $x + 1$ step by step.

Flashcard 18: Which function form is achieved by completing the square?

Answer: $(x-h)^2 + k$. This vertex form enables use of arctangent or logarithmic integrals.

Flashcard 19: How do you identify when to use long division in integration?

Answer: When the degree of the numerator is at least that of the denominator. Higher degree numerators require polynomial division before integration.

Flashcard 20: What integral technique uses $u$ substitution on quadratics?

Answer: Completing the square first. Complete the square first, then substitute $u = x + h$.

Flashcard 21: What is the integral of $\frac{x}{x^2+1}$?

Answer: $\frac{1}{2} \text{ln}|x^2+1| + C$. $u$-substitution with $u = x^2 + 1$ gives $\frac{1}{2}\ln|u|$.

Flashcard 22: How can the integral $\frac{1}{x^2+6x+13}$ be simplified?

Answer: Complete the square: $(x+3)^2 + 4$. Transforms $x^2+6x+13$ into $(x+3)^2+4$ for arctangent form.

Flashcard 23: What is the benefit of transforming a quadratic in integration?

Answer: Simplifies the integral into standard forms. Standard forms match known antiderivative formulas.

Flashcard 24: What is the first step in integrating $\frac{1}{x^2 - 4x + 5}$?

Answer: Complete the square: $(x-2)^2 + 1$. Transform to $(x-2)^2 + 1$ for arctangent integration.

Flashcard 25: Integrate $\frac{1}{(x-2)^2 + 4}$. What is the result?

Answer: $\frac{1}{2} \text{arctan} \frac{x-2}{2} + C$. Use $u = \frac{x-2}{2}$ substitution with $\frac{1}{u^2+1}$ form.

Flashcard 26: What technique helps simplify integration of rational functions?

Answer: Polynomial long division. Separates improper fractions into polynomial plus proper fraction.

Flashcard 27: When is completing the square unnecessary in integration?

Answer: When the quadratic is already a perfect square. No algebraic manipulation needed when already in standard form.

Flashcard 28: Complete the square for $x^2 - 4x + 7$. What is the result?

Answer: $(x-2)^2 + 3$. Complete the square: $(-4/2)^2 = 4$, so $7-4=3$.

Flashcard 29: When dividing $x^3 - 1$ by $x - 1$, what do you get?

Answer: $x^2 + x + 1$. Factor theorem: $(x-1)$ divides $x^3-1$ exactly.

Flashcard 30: What is the integral of $\frac{x}{x^2 + 4}$?

Answer: $\frac{1}{2} \text{ln}|x^2+4| + C$. Use $u$-substitution with $u = x^2 + 4$.