Choosing an Antidifferentiation… — AP® Calculus BC Notes

You now know every antidifferentiation technique on the BC exam: the basic rules, substitution, long division, completing the square, integration by parts and linear partial fractions. The hard part is no longer doing them; it is recognizing which one an integral wants. This page gives a reliable order of questions, then practice with the look-alikes that trip students up.

Ask the questions in order

$A flowchart of six questions in order, each with a yes arrow to a technique: basic form after rewriting? basic rules. Inside function whose derivative is a factor? substitution, with new limits. Rational with top degree at least bottom degree? long division first. Quadratic on the bottom with no real zeros? complete the square, then arctan. BC only: a product with a factor that gets simpler when differentiated? integration by parts. BC only: bottom with distinct linear factors? linear partial fractions. If none fits, rewrite and start again.$

Figure 1. Six questions, in order: the four every course asks, then the two BC adds. Most integrals are settled by the first two. LevlPrep original diagram.

Is it a basic form, or a sum of them after rewriting? Rewrite roots as powers ( $\frac{x + 1}{\sqrt{x}}$ $= x^{1/2} + x^{-1/2}$ ), expand products, and split a fraction whose bottom is a single term ( $\frac{x^3 + 2}{x} = x^2 + \frac{2}{x}$ ).
Is there an inside function whose derivative is also a factor? Then substitute: $\int x e^{x^2}\,dx$ with $u = x^2$ . A constant factor can be fixed; a missing variable factor cannot.
Is it a fraction whose top has degree at least the bottom's? Divide first: $\frac{x^2 + 1}{x - 1}$ $= x$ $+ 1$ $+ \frac{2}{x - 1}$ .
Is it a constant over a quadratic with no real zeros? Complete the square and use $\int \frac{du}{u^2 + a^2}$ $= \frac{1}{a}\arctan\frac{u}{a}$ $+ C$ .
Is it a product with one factor that gets simpler when differentiated? Use integration by parts: $\int x e^{2x}\,dx$ with $u = x$ , $dv = e^{2x}\,dx$ . A logarithm or an inverse trigonometric function goes in $u$ .
Does the bottom factor into distinct linear factors? Use linear partial fractions: $\frac{1}{x^2 - 1}$ $= \frac{1/2}{x - 1} - \frac{1/2}{x + 1}$ . Divide first if the top's degree is not below the bottom's.

Then check by differentiating. It takes ten seconds and catches sign errors and missing factors.

Look-alikes

Small changes in an integrand change the method completely. Compare:

Integral	Method	Result
$\int \frac{1}{x^2 + 1}\,dx$	basic (arctan)	$\arctan x + C$
$\int \frac{x}{x^2 + 1}\,dx$	substitution, $u = x^2 + 1$	$\frac{1}{2}\ln(x^2 + 1) + C$
$\int \frac{x^2}{x^2 + 1}\,dx$	division	$x - \arctan x + C$
$\int \frac{1}{1 + 4x^2}\,dx$	substitution, $u = 2x$	$\frac{1}{2}\arctan(2x) + C$
$\int x e^{x^2}\,dx$	substitution, $u = x^2$	$\frac{1}{2}e^{x^2} + C$
$\int e^{x^2}\,dx$	none: no elementary antiderivative	a calculator for definite integrals
$\int x e^{x}\,dx$	integration by parts, $u = x$	$(x - 1)e^x + C$
$\int \frac{1}{x^2 - 1}\,dx$	linear partial fractions	$\frac{1}{2}\ln\left\|\frac{x - 1}{x + 1}\right\| + C$

Worked example 1. Find $\int x\sqrt{x - 1}\,dx$ .

Basic? No: a product. Inside function: $x - 1$ , with derivative 1, so $u = x - 1$ and $du = dx$ . But an extra $x$ remains. Write it in terms of $u$ : $x = u + 1$ .

$\int (u + 1)u^{1/2}\,du$ $= \int \left(u^{3/2} + u^{1/2}\right)du$ $= \frac{2}{5}u^{5/2}$ $+ \frac{2}{3}u^{3/2}$ $+ C$ .

Back to $x$ : $\frac{2}{5}(x - 1)^{5/2}$ $+ \frac{2}{3}(x - 1)^{3/2}$ $+ C$ . Every $x$ must be replaced, including the one outside the root.

Worked example 2. Find $\int_1^e \frac{1 + \ln x}{x}\,dx$ .

The factor $\frac{1}{x}$ is the derivative of $\ln x$ , so let $u = 1 + \ln x$ , $du = \frac{1}{x}dx$ .

Limits: $x = 1 \to u = 1$ ; $x = e \to u = 2$ .

$\int_1^2 u\,du$ $= \frac{u^2}{2}\Big|_1^2$ $= 2 - \frac{1}{2}$ $= \frac{3}{2}$ .

Reading someone else's work

The exam sometimes shows work and asks you to find the error. Check each line against its rule: Is $du$ right? Does the constant factor match? Were the limits changed? Is the antiderivative right (differentiate it)? In $\int_0^1 \frac{6x}{x^2 + 1}\,dx$ , writing $3\int_0^1 \frac{du}{u}$ keeps the $x$ -limits; the correct line is $3\int_1^2 \frac{du}{u}$ $= 3\ln 2$ .

Try substitution and rewriting before integration by parts or partial fractions: they are quicker when they work. An infinite limit of integration, or an integrand that blows up on the interval, makes the integral improper (6.13): write it as a limit before integrating.

Choosing an Antidifferentiation Technique

Ask the questions in order

Look-alikes

Reading someone else's work