Chapter 6: Techniques of Integration
In earlier chapters, we learned the basic rules of integration: the power rule, exponential and logarithmic integrals, and trigonometric integrals. But many functions that arise in applications cannot be integrated using those basic rules alone. In this chapter, we develop more powerful techniques that allow us to evaluate a much wider class of integrals. These methods—integration by parts, partial fraction decomposition, and the handling of improper integrals—form the backbone of the AP Calculus BC exam and are highly useful even within the AB curriculum for building deeper conceptual understanding.
Compare a function f(x) (blue) with its antiderivative F(x)+C (green). Adjust C to see the family of antiderivatives.
6.1 Integration by Parts BC
Integration by parts is the integration counterpart of the product rule for differentiation. Recall that if $u$ and $v$ are differentiable functions of $x$, the product rule states:
$$\frac{d}{dx}[u \cdot v] = u \frac{dv}{dx} + v \frac{du}{dx}$$Integrating both sides and rearranging, we arrive at the integration by parts formula:
$$\int u\,dv = uv - \int v\,du$$
This transforms one integral into another that is (hopefully) easier to evaluate. The key challenge is choosing $u$ and $dv$ wisely.
The LIATE Rule
A helpful mnemonic for choosing $u$ is the LIATE rule. When the integrand is a product of two different types of functions, choose $u$ to be the function that appears first in the following list:
- L – Logarithmic functions ($\ln x$, $\log x$)
- I – Inverse trigonometric functions ($\arctan x$, $\arcsin x$)
- A – Algebraic functions ($x^2$, $x^3$, $\sqrt{x}$)
- T – Trigonometric functions ($\sin x$, $\cos x$)
- E – Exponential functions ($e^x$, $2^x$)
The idea is that functions higher in this list typically become simpler when differentiated, while functions lower in the list are easy to integrate. This is a guideline, not an absolute rule—there are exceptions—but it works remarkably well in practice for AP-level problems.
Tabular Method (Repeated Integration by Parts)
When you need to apply integration by parts multiple times in succession—typically when $u$ is a polynomial and $dv$ involves $e^x$, $\sin x$, or $\cos x$—the tabular method provides a streamlined shortcut. Set up a table with two columns: one for successive derivatives of $u$ (until you reach zero) and one for successive integrals of $dv$. Then multiply diagonally, alternating signs starting with positive ($+, -, +, -, \ldots$), and sum all the products. This avoids the tedious bookkeeping of repeated applications.
Using LIATE, choose $u = x$ (algebraic) and $dv = e^x\,dx$ (exponential).
Then $du = dx$ and $v = e^x$.
Applying the formula:
$$\int x e^x \, dx = x e^x - \int e^x \, dx = x e^x - e^x + C = e^x(x - 1) + C$$Using the tabular method with $u = x^2$ and $dv = \sin x\,dx$:
| Sign | Derivatives of $x^2$ | Integrals of $\sin x$ |
|---|---|---|
| $+$ | $x^2$ | $-\cos x$ |
| $-$ | $2x$ | $-\sin x$ |
| $+$ | $2$ | $\cos x$ |
| $0$ |
Multiplying diagonally with alternating signs:
$$\int x^2 \sin x \, dx = (+)(x^2)(-\cos x) + (-)(2x)(-\sin x) + (+)(2)(\cos x) + C$$ $$= -x^2 \cos x + 2x \sin x + 2\cos x + C$$This is a classic application where integration by parts is needed even though only one function is apparent. Write the integrand as $(\ln x)(1)$ and choose $u = \ln x$ and $dv = dx$.
Then $du = \frac{1}{x}\,dx$ and $v = x$.
$$\int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x}\,dx = x \ln x - \int 1\,dx = x \ln x - x + C$$This result is worth memorizing: $\displaystyle\int \ln x\,dx = x\ln x - x + C$.
6.2 Partial Fraction Decomposition BC
Partial fraction decomposition is a technique for integrating rational functions—that is, fractions where both the numerator and denominator are polynomials. The idea is to break a complicated fraction into a sum of simpler fractions whose integrals we already know. This technique appears frequently on the AP Calculus BC exam.
Before decomposing, ensure the rational function is proper: the degree of the numerator must be less than the degree of the denominator. If it is not, perform polynomial long division first to extract a polynomial quotient plus a proper remainder fraction.
Case 1: Distinct Linear Factors
If the denominator factors into distinct linear factors, such as $(x - a)(x - b)(x - c)$, then:
$$\frac{P(x)}{(x-a)(x-b)(x-c)} = \frac{A}{x-a} + \frac{B}{x-b} + \frac{C}{x-c}$$The constants $A$, $B$, $C$ are found by multiplying both sides by the denominator and solving the resulting system of equations (or using the "cover-up" method: set $x = a$ to find $A$, set $x = b$ to find $B$, etc.).
Case 2: Repeated Linear Factors
If a linear factor is repeated, such as $(x - a)^n$, then the decomposition includes terms for each power up to $n$:
$$\frac{P(x)}{(x-a)^3} = \frac{A}{x-a} + \frac{B}{(x-a)^2} + \frac{C}{(x-a)^3}$$The coefficients are found similarly, though you may need to combine the cover-up method with comparing coefficients for the higher-order terms.
Decompose:
$$\frac{5x+3}{(x+1)(x-2)} = \frac{A}{x+1} + \frac{B}{x-2}$$Multiply both sides by $(x+1)(x-2)$:
$$5x + 3 = A(x-2) + B(x+1)$$Cover-up method: Set $x = 2$: $\;5(2)+3 = B(3) \implies 13 = 3B \implies B = \frac{13}{3}$.
Set $x = -1$: $\;5(-1)+3 = A(-3) \implies -2 = -3A \implies A = \frac{2}{3}$.
Therefore:
$$\int \frac{5x+3}{(x+1)(x-2)}\,dx = \frac{2}{3}\int\frac{dx}{x+1} + \frac{13}{3}\int\frac{dx}{x-2}$$ $$= \frac{2}{3}\ln|x+1| + \frac{13}{3}\ln|x-2| + C$$This involves a repeated linear factor. Decompose:
$$\frac{3x+5}{(x+1)^2} = \frac{A}{x+1} + \frac{B}{(x+1)^2}$$Multiply both sides by $(x+1)^2$:
$$3x + 5 = A(x+1) + B$$Set $x = -1$: $\;3(-1)+5 = B \implies B = 2$.
Compare coefficients of $x$: $3 = A$.
Therefore:
$$\int \frac{3x+5}{(x+1)^2}\,dx = 3\int\frac{dx}{x+1} + 2\int(x+1)^{-2}\,dx$$ $$= 3\ln|x+1| - \frac{2}{x+1} + C$$6.3 Improper Integrals BC
So far, our definite integrals have always had finite limits of integration and continuous integrands. An improper integral extends the concept to cases where one or both of these conditions fail. Understanding improper integrals is essential for BC students, as they appear in contexts ranging from probability to series convergence tests.
Type 1: Infinite Limits of Integration
If $f$ is continuous on $[a, \infty)$, we define: $$\int_a^{\infty} f(x)\,dx = \lim_{t \to \infty} \int_a^{t} f(x)\,dx$$ If the limit exists and is finite, the integral converges. Otherwise, it diverges. Similarly for integrals from $-\infty$ to $b$, or from $-\infty$ to $\infty$ (split at any convenient point).
Type 2: Discontinuous Integrands
If $f$ has a vertical asymptote at $x = c$ within or at an endpoint of the interval $[a,b]$, then we treat it as a limit as well. For example, if $f$ is continuous on $[a, c)$ and has a vertical asymptote at $x = c$:
$$\int_a^{c} f(x)\,dx = \lim_{t \to c^-} \int_a^{t} f(x)\,dx$$The Comparison Test for Improper Integrals
Suppose $0 \le f(x) \le g(x)$ for all $x \ge a$. Then:
- If $\displaystyle\int_a^{\infty} g(x)\,dx$ converges, then $\displaystyle\int_a^{\infty} f(x)\,dx$ also converges.
- If $\displaystyle\int_a^{\infty} f(x)\,dx$ diverges, then $\displaystyle\int_a^{\infty} g(x)\,dx$ also diverges.
A key reference integral for comparison is the $p$-integral: $\displaystyle\int_1^{\infty} \frac{1}{x^p}\,dx$ converges if and only if $p > 1$. When $p > 1$, its value is $\frac{1}{p-1}$.
Improper integral convergence: ∫1∞ 1/xp dx converges when p > 1 and diverges when p ≤ 1. Adjust p and b to explore.
Replace the infinite upper limit with a variable $t$ and take the limit:
$$\int_1^{\infty} \frac{1}{x^2}\,dx = \lim_{t\to\infty}\int_1^{t} x^{-2}\,dx = \lim_{t\to\infty}\left[-\frac{1}{x}\right]_1^{t} = \lim_{t\to\infty}\left(-\frac{1}{t} + 1\right) = 0 + 1 = 1$$The integral converges to $1$. This is consistent with the $p$-integral rule since $p = 2 > 1$.
The integrand $\frac{1}{\sqrt{x}}$ has a vertical asymptote at $x = 0$ (Type 2 improper integral).
$$\int_0^{1}\frac{1}{\sqrt{x}}\,dx = \lim_{t\to 0^+}\int_t^{1} x^{-1/2}\,dx = \lim_{t\to 0^+}\left[2\sqrt{x}\right]_t^{1} = \lim_{t\to 0^+}\left(2 - 2\sqrt{t}\right) = 2 - 0 = 2$$The integral converges to $2$, even though the integrand blows up at $x = 0$. The area under the curve near the asymptote grows slowly enough to remain finite.
6.4 Integration Using Tables and Technology AB/BC
Not every integral yields to elementary techniques. In practice—and on portions of the AP exam that allow a calculator—it is important to recognize when an integral matches a standard form from a table, when numerical methods (like a Riemann sum or the Trapezoidal Rule) are appropriate, and when technology can assist.
Recognizing Standard Forms
Many integrals can be evaluated by recognizing that the integrand fits a known pattern, perhaps after a substitution. Some essential standard forms include:
- $\displaystyle\int \frac{du}{u^2 + a^2} = \frac{1}{a}\arctan\!\left(\frac{u}{a}\right) + C$
- $\displaystyle\int \frac{du}{\sqrt{a^2 - u^2}} = \arcsin\!\left(\frac{u}{a}\right) + C$
- $\displaystyle\int \frac{du}{u\sqrt{u^2 - a^2}} = \frac{1}{a}\text{arcsec}\!\left(\frac{|u|}{a}\right) + C$
- $\displaystyle\int e^{au}\sin(bu)\,du = \frac{e^{au}}{a^2+b^2}\bigl(a\sin(bu) - b\cos(bu)\bigr) + C$
Decision Flowchart: Which Technique to Use
When facing an integral, work through these questions in order:
- Is it a basic form? Check if a direct antiderivative rule applies (power rule, exponential, trig).
- Can $u$-substitution simplify it? Look for a composite function whose inner function's derivative appears in the integrand.
- Is it a product of two different types of functions? Try integration by parts (LIATE).
- Is it a rational function? If the integrand is a ratio of polynomials, use partial fractions.
- Does it match a standard form? Try completing the square or a trigonometric substitution to match one of the table entries above.
- None of the above? Use technology (CAS or numerical integration) on the calculator-active portion of the exam.
Complete the square in the denominator:
$$x^2 + 4x + 13 = (x^2 + 4x + 4) + 9 = (x+2)^2 + 3^2$$Now this matches the arctangent standard form with $u = x+2$ and $a = 3$:
$$\int \frac{dx}{(x+2)^2 + 9} = \frac{1}{3}\arctan\!\left(\frac{x+2}{3}\right) + C$$Split the integrand into two parts:
$$\int \frac{2x+1}{x^2+1}\,dx = \int\frac{2x}{x^2+1}\,dx + \int\frac{1}{x^2+1}\,dx$$For the first integral, use $u$-substitution with $u = x^2+1$, $du = 2x\,dx$:
$$\int\frac{2x}{x^2+1}\,dx = \ln|x^2+1| = \ln(x^2+1)$$(since $x^2 + 1 > 0$ always). The second integral is a standard arctangent form:
$$\int\frac{1}{x^2+1}\,dx = \arctan x$$Combined result:
$$\int \frac{2x+1}{x^2+1}\,dx = \ln(x^2+1) + \arctan x + C$$6.5 Practice Problems
Test your understanding of the techniques covered in this chapter. Each problem targets one or more of the methods from Sections 6.1–6.4. Try to solve each problem before revealing the solution.
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First application: $u = \sin x$, $dv = e^x dx$. Then $du = \cos x\,dx$, $v = e^x$. $$I = e^x \sin x - \int e^x \cos x\,dx$$ Second application: $u = \cos x$, $dv = e^x dx$. Then $du = -\sin x\,dx$, $v = e^x$. $$I = e^x \sin x - \left(e^x \cos x - \int e^x(-\sin x)\,dx\right) = e^x\sin x - e^x\cos x - I$$ Solving for $I$: $$2I = e^x(\sin x - \cos x) \implies I = \frac{e^x(\sin x - \cos x)}{2} + C$$
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First, find the antiderivative using IBP: $u = x$, $dv = e^{-x}dx$. Then $du = dx$, $v = -e^{-x}$. $$\int x e^{-x}\,dx = -xe^{-x} + \int e^{-x}\,dx = -xe^{-x} - e^{-x} = -e^{-x}(x+1)$$ Now evaluate the improper integral: $$\int_0^{\infty}xe^{-x}\,dx = \lim_{t\to\infty}\left[-e^{-x}(x+1)\right]_0^t = \lim_{t\to\infty}\left(-\frac{t+1}{e^t} + 1\right)$$ By L'Hopital's Rule (or by recognizing that exponentials dominate polynomials), $\lim_{t\to\infty}\frac{t+1}{e^t} = 0$. $$\int_0^{\infty}xe^{-x}\,dx = 0 + 1 = 1$$ This result has a beautiful connection to probability: it is the mean of the exponential distribution with rate parameter $\lambda = 1$.