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AP Calculus BC Flashcards: Selecting Techniques For Antidifferentiation

Study Selecting Techniques For Antidifferentiation in AP Calculus BC with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

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What this deck covers

This deck focuses on Selecting Techniques For Antidifferentiation, 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.

AP Calculus BC Flashcards: Selecting Techniques For Antidifferentiation

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QUESTION

Which technique is used for xex2xe^{x^2}xex2?

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ANSWER

U-Substitution. Let u=x2u = x^2u=x2, then du=2xdxdu = 2x dxdu=2xdx for substitution.

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Flashcard 1: Which technique is used for xex2xe^{x^2}xex2?

Answer: U-Substitution. Let u=x2u = x^2u=x2, then du=2xdxdu = 2x dxdu=2xdx for substitution.

Flashcard 2: What is the antiderivative of xnx^nxn for n≠−1n \neq -1n=−1?

Answer: xn+1n+1+C\frac{x^{n+1}}{n+1} + Cn+1xn+1​+C. Power rule for integration: increase exponent by 1, divide by new exponent.

Flashcard 3: Which method is used for x2exx^2 e^xx2ex?

Answer: Integration by Parts. Product x2exx^2 e^xx2ex requires repeated integration by parts.

Flashcard 4: What is the antiderivative of exe^xex?

Answer: ex+Ce^x + Cex+C. The derivative of exe^xex is itself, so integration reverses this.

Flashcard 5: What is the antiderivative of eaxe^{ax}eax?

Answer: 1aeax+C\frac{1}{a}e^{ax} + Ca1​eax+C. Chain rule in reverse: derivative of 1aeax\frac{1}{a}e^{ax}a1​eax is eaxe^{ax}eax.

Flashcard 6: Which method is used for x2exx^2 e^xx2ex?

Answer: Integration by Parts. Product x2exx^2 e^xx2ex requires repeated integration by parts.

Flashcard 7: Which technique is used for xex2xe^{x^2}xex2?

Answer: U-Substitution. Let u=x2u = x^2u=x2, then du=2xdxdu = 2x dxdu=2xdx for substitution.

Flashcard 8: What is the antiderivative of xnx^nxn for n≠−1n \neq -1n=−1?

Answer: xn+1n+1+C\frac{x^{n+1}}{n+1} + Cn+1xn+1​+C. Power rule for integration: increase exponent by 1, divide by new exponent.

Flashcard 9: Which technique is most appropriate for ∫x2−16 dx\int \sqrt{x^2-16}\,dx∫x2−16​dx?

Answer: Use trig substitution: x=4sec⁡(θ)x=4\sec(\theta)x=4sec(θ). Square root of x2−a2x^2-a^2x2−a2 form requires secant substitution.

Flashcard 10: Which technique is most appropriate for ∫x5 dx\int x^5\,dx∫x5dx?

Answer: Use the power rule (basic antiderivative). Polynomial integrands use the power rule directly.

Flashcard 11: Which technique is most appropriate for ∫e3x dx\int e^{3x}\,dx∫e3xdx?

Answer: Use reverse chain rule: ∫eaxdx=1aeax+C\int e^{ax}dx=\frac{1}{a}e^{ax}+C∫eaxdx=a1​eax+C. The coefficient aaa in eaxe^{ax}eax moves to the denominator.

Flashcard 12: Which technique is most appropriate for ∫cos⁡(7x) dx\int \cos(7x)\,dx∫cos(7x)dx?

Answer: Use reverse chain rule: ∫cos⁡(ax)dx=1asin⁡(ax)+C\int \cos(ax)dx=\frac{1}{a}\sin(ax)+C∫cos(ax)dx=a1​sin(ax)+C. The coefficient aaa in cos⁡(ax)\cos(ax)cos(ax) moves to the denominator.

Flashcard 13: Which technique is most appropriate for ∫1x dx\int \frac{1}{x}\,dx∫x1​dx?

Answer: Use the logarithm rule: ∫1xdx=ln⁡∣x∣+C\int \frac{1}{x}dx=\ln|x|+C∫x1​dx=ln∣x∣+C. The integral of 1x\frac{1}{x}x1​ is the natural logarithm.

Flashcard 14: Which technique is most appropriate for ∫12x−5 dx\int \frac{1}{2x-5}\,dx∫2x−51​dx?

Answer: Use reverse chain rule: 12ln⁡∣2x−5∣+C\frac{1}{2}\ln|2x-5|+C21​ln∣2x−5∣+C. Linear denominator requires chain rule adjustment by 12\frac{1}{2}21​.

Flashcard 15: Which technique is most appropriate for ∫(x2+1)7⋅2x dx\int (x^2+1)^7\cdot 2x\,dx∫(x2+1)7⋅2xdx?

Answer: Use uuu-substitution with u=x2+1u=x^2+1u=x2+1. The derivative of u=x2+1u=x^2+1u=x2+1 is 2x2x2x, which appears in the integrand.

Flashcard 16: Which technique is most appropriate for ∫xx2+9 dx\int \frac{x}{x^2+9}\,dx∫x2+9x​dx?

Answer: Use uuu-substitution with u=x2+9u=x^2+9u=x2+9. The numerator xxx is half the derivative of the denominator.

Flashcard 17: Which technique is most appropriate for ∫xx2+4 dx\int x\sqrt{x^2+4}\,dx∫xx2+4​dx?

Answer: Use uuu-substitution with u=x2+4u=x^2+4u=x2+4. The xxx factor is half the derivative of what's under the radical.

Flashcard 18: Which technique is most appropriate for ∫xex2 dx\int x e^{x^2}\,dx∫xex2dx?

Answer: Use uuu-substitution with u=x2u=x^2u=x2. The xxx factor is half the derivative of the exponent.

Flashcard 19: Which technique is most appropriate for ∫xln⁡(x) dx\int x\ln(x)\,dx∫xln(x)dx?

Answer: Use integration by parts. Products of polynomials and logarithms require parts.

Flashcard 20: Which technique is most appropriate for ∫xsin⁡(x) dx\int x\sin(x)\,dx∫xsin(x)dx?

Answer: Use integration by parts. Products of polynomials and trig functions require parts.

Flashcard 21: Which technique is most appropriate for ∫excos⁡(x) dx\int e^x\cos(x)\,dx∫excos(x)dx?

Answer: Use integration by parts twice (tabular/recursive). Products of exponentials and trig functions need repeated parts.

Flashcard 22: Which technique is most appropriate for ∫x2+1x−3 dx\int \frac{x^2+1}{x-3}\,dx∫x−3x2+1​dx?

Answer: Use algebraic division, then integrate term-by-term. Improper rational function requires polynomial division first.

Flashcard 23: Which technique is most appropriate for ∫1x2−9 dx\int \frac{1}{x^2-9}\,dx∫x2−91​dx?

Answer: Use partial fractions (factor x2−9=(x−3)(x+3)x^2-9=(x-3)(x+3)x2−9=(x−3)(x+3)). Factorable quadratic denominator uses partial fractions.

Flashcard 24: Which technique is most appropriate for ∫1x2+9 dx\int \frac{1}{x^2+9}\,dx∫x2+91​dx?

Answer: Use arctangent form: ∫1x2+a2dx=1aarctan⁡(xa)+C\int \frac{1}{x^2+a^2}dx=\frac{1}{a}\arctan(\frac{x}{a})+C∫x2+a21​dx=a1​arctan(ax​)+C. Matches the arctangent integral form with a=3a=3a=3.

Flashcard 25: Which technique is most appropriate for ∫19−x2 dx\int \frac{1}{\sqrt{9-x^2}}\,dx∫9−x2​1​dx?

Answer: Use arcsine form: ∫1a2−x2dx=arcsin⁡(xa)+C\int \frac{1}{\sqrt{a^2-x^2}}dx=\arcsin(\frac{x}{a})+C∫a2−x2​1​dx=arcsin(ax​)+C. Matches the arcsine integral form with a=3a=3a=3.

Flashcard 26: Which technique is most appropriate for ∫9−x2 dx\int \sqrt{9-x^2}\,dx∫9−x2​dx?

Answer: Use trig substitution: x=3sin⁡(θ)x=3\sin(\theta)x=3sin(θ). Square root of a2−x2a^2-x^2a2−x2 form requires sine substitution.

Flashcard 27: Which technique is most appropriate for ∫x2+16 dx\int \sqrt{x^2+16}\,dx∫x2+16​dx?

Answer: Use trig substitution: x=4tan⁡(θ)x=4\tan(\theta)x=4tan(θ). Square root of x2+a2x^2+a^2x2+a2 form requires tangent substitution.

Flashcard 28: Which technique is most appropriate for ∫sin⁡3(x)cos⁡2(x) dx\int \sin^3(x)\cos^2(x)\,dx∫sin3(x)cos2(x)dx?

Answer: Use trig identities; save a sin⁡(x)\sin(x)sin(x) and substitute u=cos⁡(x)u=\cos(x)u=cos(x). Odd power of sine allows uuu-substitution after saving one sine.

Flashcard 29: What is the antiderivative of eaxe^{ax}eax?

Answer: 1aeax+C\frac{1}{a}e^{ax} + Ca1​eax+C. Chain rule in reverse: derivative of 1aeax\frac{1}{a}e^{ax}a1​eax is eaxe^{ax}eax.

Flashcard 30: What is the antiderivative of exe^xex?

Answer: ex+Ce^x + Cex+C. The derivative of exe^xex is itself, so integration reverses this.