Choosing an Integration Technique: The Ordered Decision Checklist

To know which integration technique to use, run an ordered checklist and stop at the first match: first, is it a basic antiderivative or one small rewrite away (split the fraction, expand, or use an identity), then just do it. If not, is there an inside function whose derivative is also present? That is u-substitution, and you should try it first because it is the most common. If it is a product of two unlike functions like a polynomial times e to the x or x times sine x, use integration by parts and let LIATE pick u. If it is a polynomial over a factorable polynomial, use partial fractions; and if you see a root of a squared plus or minus x squared, use trig substitution. Name the technique before you integrate — that decision is most of the battle.

Integration feels like guessing because nobody hands you a decision rule — you were taught the techniques but not how to CHOOSE one. Here is the rule. Before you integrate anything, run this ordered checklist and stop at the first match. Naming the technique is 80% of the problem; the algebra after is the easy part.

1. Is it a basic antiderivative, or one tiny rewrite away? If you already know the answer (powers, e^x, sin, cos, 1/x), just do it directly. Always try to simplify first: split a fraction into separate terms, expand a product, or use a trig identity. Many "hard" integrals collapse the moment you rewrite them.

2. Is there an inside function whose derivative is also sitting in the integrand? That is u-substitution — try this FIRST, because it is by far the most common technique. Look for a composite piece (something inside parentheses, an exponent, or under a root) whose derivative appears as a factor, off by at most a constant.

3. Is it a PRODUCT of two unlike functions? Polynomial times e^x, x times sin x, or a lone ln x — that is integration by parts. Use LIATE (Logs, Inverse trig, Algebraic, Trig, Exponential) to pick which factor is u.

4. Is it a rational function — a polynomial divided by a polynomial — with a factorable denominator? That is partial fractions: break the single fraction into a sum of simpler ones you can integrate term by term.

5. Do you see a root of the form a^2 minus x^2, a^2 plus x^2, or x^2 minus a^2? That is trig substitution.

The first three cover the vast majority of what you will see. When in doubt, rewrite, then reach for u-substitution.

u = \text{inside function} \;\Rightarrow\; du = u'(x)\,dx

How do you decide which integration technique to use?

\int x\,e^{x^2}\,dx \;\xrightarrow{\;u=x^2,\;du=2x\,dx\;}\; \tfrac{1}{2}\int e^{u}\,du = \tfrac{1}{2}e^{x^2}+C

Worked example: naming the technique for the integral of x times e to the x squared

Name the technique, then evaluate the integral of x·e^(x²) dx.

First name the technique — that is the whole game here. Scan the checklist: it is not a basic antiderivative, but there IS an inside function, x squared, sitting in the exponent, and its derivative 2x appears as a factor (we have an x out front, off only by the constant 2). That is the u-substitution signature, so we use u-sub.

Let u be the inside function and differentiate:

u = x^2 \quad\Rightarrow\quad du = 2x\,dx \quad\Rightarrow\quad x\,dx = \tfrac{1}{2}\,du

Substitute so the whole integral is written in terms of u:

\int x\,e^{x^2}\,dx = \int e^{u}\cdot\tfrac{1}{2}\,du = \tfrac{1}{2}\int e^{u}\,du

Integrate the clean, basic form and substitute x squared back for u:

\tfrac{1}{2}\int e^{u}\,du = \tfrac{1}{2}e^{u}+C = \tfrac{1}{2}e^{x^2}+C

Verify by differentiating the answer — the chain rule should hand you back the original integrand:

\frac{d}{dx}\left[\tfrac{1}{2}e^{x^2}\right] = \tfrac{1}{2}\cdot e^{x^2}\cdot 2x = x\,e^{x^2} \checkmark

Another worked example

Name the technique for the integral of 2x times (x squared plus 1) to the 5th. Step 1: run the checklist — the inside function is x squared plus 1, and its derivative 2x is sitting right there as a factor, so this is u-substitution. Step 2: let u be x squared plus 1, so du is 2x dx, and the integral becomes the integral of u to the 5th du. Step 3: integrate to get u to the 6th over 6, then substitute back to get (x squared plus 1) to the 6th over 6, plus C.

Aligned with the College Board AP Calculus CED — Unit 6 (Integration and Accumulation of Change), Topics 6.9 (Integrating Using Substitution), 6.10 (Long Division and Completing the Square), and 6.14 (Selecting Techniques for Antidifferentiation), plus the integration-by-parts and partial-fractions techniques of BC Topics 6.11-6.12. College Board ↗

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