Choosing an Antidifferentiation… — AP® Calculus AB Notes

You now know every antidifferentiation technique on the AB exam: the basic rules, substitution, long division and completing the square. 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

Figure 1. Four questions, in order. 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 $\int \frac{du}{u^2 + a^2}$ $= \frac{1}{a}\arctan\frac{u}{a}$ $+ C$ .

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

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 $3\int_1^2 \frac{du}{u}$ $= 3\ln 2$ .

Integrals that need integration by parts, partial fractions or an infinite limit of integration belong to the BC course. If an AB integral seems to need one of those, look again for a substitution or a rewrite, or check whether it is a calculator question.