AP Calculus BC

Advanced Integration Techniques

Calculus BC

✓By the end of this lesson students will be able to apply the technique of integration by parts to evaluate definite and indefinite integrals.
✓By the end of this lesson students will be able to decompose rational functions into partial fractions and use this decomposition to evaluate integrals.
✓By the end of this lesson students will be able to evaluate improper integrals with infinite limits of integration or infinite discontinuities within the interval.
✓By the end of this lesson students will be able to determine whether an improper integral converges or diverges.

Key concepts

Integration by Parts

Integration by parts is a technique used to integrate products of functions. It is derived from the product rule for differentiation. The key is to choose 'u' and 'dv' such that the new integral, ∫v du, is easier to evaluate than the original integral. A common heuristic for choosing 'u' is LIATE: Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential. The function appearing earliest in this list should generally be chosen as 'u'.

∫u dv = uv - ∫v du

Partial Fractions

The method of partial fractions is used to integrate rational functions (functions that are ratios of polynomials). If the degree of the numerator is greater than or equal to the degree of the denominator, polynomial long division should be performed first. Then, the rational function is decomposed into a sum of simpler fractions whose denominators are the factors of the original denominator. These simpler fractions are easier to integrate.

For a rational function P(x)/Q(x), where Q(x) can be factored into linear and/or irreducible quadratic factors:\n- For each distinct linear factor (ax+b): A/(ax+b)\n- For each repeated linear factor (ax+b)^n: A1/(ax+b) + A2/(ax+b)^2 + ... + An/(ax+b)^n\n- For each distinct irreducible quadratic factor (ax^2+bx+c): (Ax+B)/(ax^2+bx+c)\n- For each repeated irreducible quadratic factor (ax^2+bx+c)^n: (A1x+B1)/(ax^2+bx+c) + ... + (Anx+Bn)/(ax^2+bx+c)^n

Improper Integrals

Improper integrals are definite integrals where either one or both of the limits of integration are infinite, or the integrand has an infinite discontinuity (a vertical asymptote) at one or more points within the interval of integration. These integrals are evaluated using limits. If the limit exists and is finite, the integral converges; otherwise, it diverges.

Type 1 (Infinite Limits):\n- ∫a^∞ f(x) dx = lim(b→∞) ∫a^b f(x) dx\n- ∫-∞^b f(x) dx = lim(a→-∞) ∫a^b f(x) dx\n- ∫-∞^∞ f(x) dx = ∫-∞^c f(x) dx + ∫c^∞ f(x) dx (where c is any real number)\n\nType 2 (Infinite Discontinuities):\n- If f has an infinite discontinuity at b: ∫a^b f(x) dx = lim(c→b-) ∫a^c f(x) dx\n- If f has an infinite discontinuity at a: ∫a^b f(x) dx = lim(c→a+) ∫c^b f(x) dx\n- If f has an infinite discontinuity at c, where a < c < b: ∫a^b f(x) dx = ∫a^c f(x) dx + ∫c^b f(x) dx

Key facts to remember

1Integration by Parts: ∫u dv = uv - ∫v du. Use LIATE to choose 'u' (Logarithmic, Inverse Trig, Algebraic, Trig, Exponential).
2Partial Fractions: Decompose rational functions into simpler fractions. Ensure the degree of the numerator is less than the degree of the denominator; if not, perform polynomial long division first.
3Improper Integrals: Evaluate using limits. If the limit is finite, the integral converges; otherwise, it diverges.
4Type 1 Improper Integrals: Involve infinite limits of integration (e.g., ∫a^∞ f(x) dx).
5Type 2 Improper Integrals: Involve infinite discontinuities within the interval of integration (e.g., ∫a^b f(x) dx where f(x) has a vertical asymptote at a or b, or between a and b).
6The p-series test for ∫1^∞ (1/x^p) dx: Converges if p > 1, diverges if p ≤ 1.
7The p-series test for ∫0^1 (1/x^p) dx: Converges if p < 1, diverges if p ≥ 1.

Worked examples

Example 1

Evaluate the indefinite integral: ∫x e^x dx

IIdentify u and dv. Using LIATE, 'x' is Algebraic and 'e^x' is Exponential. So, let u = x and dv = e^x dx.

IIFind du and v. Differentiate u to get du = dx. Integrate dv to get v = e^x.

IIIApply the integration by parts formula: ∫u dv = uv - ∫v du.

IVSubstitute the values: ∫x e^x dx = x e^x - ∫e^x dx.

VEvaluate the remaining integral: ∫e^x dx = e^x.

VICombine the terms and add the constant of integration: x e^x - e^x + C.

Answer

x e^x - e^x + C

Remember to add the constant of integration 'C' for indefinite integrals.

Example 2

Evaluate the indefinite integral: ∫(x+1)/(x^2+x-2) dx

IFactor the denominator: x^2 + x - 2 = (x+2)(x-1).

IISet up the partial fraction decomposition: (x+1)/((x+2)(x-1)) = A/(x+2) + B/(x-1).

IIIMultiply both sides by the common denominator (x+2)(x-1): x+1 = A(x-1) + B(x+2).

IVSolve for A and B. \n - Let x = 1: 1+1 = A(1-1) + B(1+2) => 2 = 3B => B = 2/3.\n - Let x = -2: -2+1 = A(-2-1) + B(-2+2) => -1 = -3A => A = 1/3.

VRewrite the integral using the partial fractions: ∫(1/3)/(x+2) dx + ∫(2/3)/(x-1) dx.

VIIntegrate each term: (1/3)ln|x+2| + (2/3)ln|x-1| + C.

Answer

(1/3)ln|x+2| + (2/3)ln|x-1| + C

Always factor the denominator completely before setting up the partial fractions. If the numerator's degree is greater than or equal to the denominator's, perform polynomial long division first.

Example 3

Determine if the improper integral ∫1^∞ (1/x^2) dx converges or diverges. If it converges, find its value.

IRewrite the improper integral as a limit: lim(b→∞) ∫1^b (1/x^2) dx.

IIRewrite the integrand with a negative exponent for easier integration: lim(b→∞) ∫1^b x^-2 dx.

IIIFind the antiderivative of x^-2: -x^-1 = -1/x.

IVEvaluate the definite integral: [-1/x] from 1 to b = (-1/b) - (-1/1) = 1 - 1/b.

VEvaluate the limit: lim(b→∞) (1 - 1/b).

VIAs b→∞, 1/b → 0. So, the limit is 1 - 0 = 1.

Answer

The integral converges to 1.

For improper integrals, always use limit notation correctly. The p-series test states that ∫1^∞ (1/x^p) dx converges if p > 1 and diverges if p ≤ 1. Here, p=2, so it converges.

Example 4

Determine if the improper integral ∫0^1 (1/√x) dx converges or diverges. If it converges, find its value.

IIdentify the discontinuity: The integrand 1/√x has an infinite discontinuity at x=0, which is a limit of integration.

IIRewrite the improper integral as a limit: lim(a→0+) ∫a^1 (1/√x) dx.

IIIRewrite the integrand with a negative exponent for easier integration: lim(a→0+) ∫a^1 x^(-1/2) dx.

IVFind the antiderivative of x^(-1/2): 2x^(1/2) = 2√x.

VEvaluate the definite integral: [2√x] from a to 1 = 2√1 - 2√a = 2 - 2√a.

VIEvaluate the limit: lim(a→0+) (2 - 2√a).

VIIAs a→0+, √a → 0. So, the limit is 2 - 2(0) = 2.

Answer

The integral converges to 2.

This is a Type 2 improper integral. The p-series test for ∫0^1 (1/x^p) dx states it converges if p < 1 and diverges if p ≥ 1. Here, p=1/2, so it converges.

Common mistakes

✗Incorrectly choosing 'u' and 'dv' in integration by parts, leading to a more complex integral.
✗Algebraic errors when solving for the constants (A, B, C, etc.) in partial fraction decomposition.
✗Forgetting to use limit notation when evaluating improper integrals, or incorrectly evaluating the limit.
✗Not checking for infinite discontinuities within the interval of integration for Type 2 improper integrals.
✗Incorrectly applying the p-series test or other comparison tests for improper integrals.

Exam tips

★For integration by parts, practice the LIATE rule and be prepared for integrals that require repeated application or solving for the integral (cyclic integrals).
★Master algebraic manipulation for partial fractions, including solving systems of equations for the constants. Be meticulous with signs.
★Always write out the limit notation explicitly for improper integrals. This is crucial for showing your work and earning full credit.
★When evaluating improper integrals, clearly state whether the integral converges or diverges and, if it converges, state its value.

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