Integration Strategy
Textbook Reference
| Primary source | OpenStax Calculus Volume 2, Section 3.5: “Other Strategies for Integration” |
| Direct link | https://openstax.org/books/calculus-volume-2/pages/3-5-other-strategies-for-integration |
| Textbook used in class | Stewart, Calculus, Section 7.5: “Strategy for Integration” |
Opening Scenario
With six major techniques in hand -- basic formulas, substitution, integration by parts, trigonometric integrals, trigonometric substitution, and partial fractions -- the challenge shifts from learning each technique to knowing which one to apply. Most integrands do not arrive labeled. This lesson organizes the decision process into a sequence of questions you ask about the integrand’s structure.
Quick Reference
Decision sequence:
- Can you simplify the integrand algebraically? Expand, factor, split fractions, use a trig identity, or rewrite radicals first.
- Is it a basic formula? Check the table: $x^n$, $e^x$, $\sin x$, $\cos x$, $\frac{1}{x}$, $\sec^2 x$, $\frac{1}{1+x^2}$, etc.
- Is there an obvious substitution? Look for a composite function where the derivative of the inner function appears.
- Is the integrand a product? Consider integration by parts. Try $u = $ the “harder to integrate” factor and $dv = $ the simpler one.
- Does the integrand involve $\sqrt{a^2-x^2}$, $\sqrt{a^2+x^2}$, or $\sqrt{x^2-a^2}$? Use trigonometric substitution.
- Is the integrand a rational function? Try partial fractions (after long division if improper).
- Can you convert to a known form via a trig identity? Powers of sine/cosine/secant/tangent often simplify via Pythagorean or half-angle identities.
If none of these work immediately, try combining two techniques or look for an algebraic manipulation that reveals the structure.
Key Concepts
1. Simplify First
Before reaching for a technique, check whether the integrand can be simplified. Expanding $(x+1)^2$ before integrating avoids substitution entirely. Rewriting $\dfrac{x^2+1}{x}$ as $x + x^{-1}$ makes the power rule apply directly. Using $\sin^2 x = \frac{1-\cos(2x)}{2}$ turns a tricky integral into one you know.
2. Substitution Is the First Reflex
After simplification, most students find substitution is the first technique to try because it handles a wide range of composed expressions. If you see $f(g(x))$ and $g'(x)$ (or a constant multiple) is present, substitution will work.
3. Integration by Parts for Products
When you have a product of two functions and neither is the derivative of the other, integration by parts is the natural next step. The LIATE mnemonic (Logarithm, Inverse trig, Algebraic, Trigonometric, Exponential) suggests which factor to label $u$: choose the type that appears earlier in the list.
4. Rational Functions Always Yield to Partial Fractions
If the integrand is a ratio of two polynomials, partial fractions (with possible long division) always reduces it to a sum of simpler fractions, each integrable. The process may be lengthy, but it is systematic and guaranteed to terminate.
5. Sometimes Two Techniques in Sequence
Many integrals require one technique to set up and a second to finish:
- A trig substitution may produce a trig integral requiring the sin/cos power techniques.
- Integration by parts may produce an integral that requires substitution.
- Substitution may produce a rational function requiring partial fractions.
Expect to chain techniques.
“If the first technique I try does not immediately simplify the integral, it must be wrong.” Integration is not always one step. If a substitution simplifies the integrand but does not produce a standard form, apply a second technique to the result. A failed simplification often still moved the integral closer to solvable.
Worked Example
Classify and describe the approach for each integral. Do not compute.
(a) $\displaystyle\int x^3\ln x\,dx$
Recognition: Product of an algebraic function $x^3$ and a logarithm $\ln x$. Substitution does not obviously apply (derivative of $\ln x$ is $1/x$, not $x^3$). Integration by parts with $u = \ln x$, $dv = x^3\,dx$.
(b) $\displaystyle\int\frac{x+1}{x^2-x-6}\,dx$
Recognition: Proper rational function (degree 1 numerator, degree 2 denominator). Factor: $x^2-x-6 = (x-3)(x+2)$. Partial fractions: $\dfrac{A}{x-3} + \dfrac{B}{x+2}$.
(c) $\displaystyle\int\frac{x}{\sqrt{4-x^2}}\,dx$
Recognition: Denominator contains $\sqrt{a^2-x^2}$. However, the extra factor $x$ in the numerator is (up to a constant) the derivative of $4-x^2$. Substitution $u = 4-x^2$ may be faster than trig substitution. Let $u = 4-x^2$, $du = -2x\,dx$.
(d) $\displaystyle\int e^x\sin x\,dx$
Recognition: Product of exponential and trig function. Integration by parts, applied twice. The integral appears on both sides of the equation; solve algebraically.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Jumping to integration by parts when substitution works | Using parts for $\int xe^{x^2}\,dx$ | Recognize $x = \frac{1}{2}(2x)$ and the derivative of $x^2$ is $2x$: substitution $u=x^2$ works |
| Trying partial fractions on a non-rational function | Attempting partial fractions on $\int\sqrt{x^2+1}\,dx$ | Partial fractions applies to rational functions only; $\sqrt{x^2+1}$ requires trig substitution |
| Giving up after one failed attempt | Abandoning an integral after substitution “does not work” | Simplify, then substitute; if that fails, try parts; if that fails, try trig identity or substitution on the result |
Leveled Practice
Level 1 -- Identify the Technique
Problem 1. For each integral, state the most appropriate first technique and briefly justify.
(a) $\displaystyle\int\frac{\cos x}{1+\sin^2 x}\,dx$
(b) $\displaystyle\int x^2 e^x\,dx$
(c) $\displaystyle\int\frac{3x^2-2}{x^3-x}\,dx$
Show answer
(a) Let $u = \sin x$, $du = \cos x\,dx$: integral becomes $\int\frac{du}{1+u^2} = \arctan u + C = \arctan(\sin x) + C$. Technique: substitution.
(b) Product of algebraic ($x^2$) and exponential ($e^x$): LIATE says $u = x^2$. Will need parts twice. Technique: integration by parts.
(c) Factor denominator: $x^3-x = x(x-1)(x+1)$. Rational function with distinct linear factors. Technique: partial fractions.
Level 2 -- Two-Step Problem
Problem 2. Outline (without computing) the two-step approach for $\displaystyle\int\frac{\ln x}{\sqrt{x}}\,dx$.
Show answer
Step 1: The factor $\frac{1}{\sqrt{x}} = x^{-1/2}$ integrates nicely. Use integration by parts with $u = \ln x$ and $dv = x^{-1/2}\,dx$. This gives $v = 2\sqrt{x}$ and $du = \frac{1}{x}\,dx$.
After parts: $\int\frac{\ln x}{\sqrt{x}}\,dx = 2\sqrt{x}\ln x - \int\frac{2\sqrt{x}}{x}\,dx = 2\sqrt{x}\ln x - 2\int x^{-1/2}\,dx$.
Step 2: The remaining integral $2\int x^{-1/2}\,dx = 4\sqrt{x}$ uses the power rule.
Mastery Checklist
Mental Model
Choosing an integration technique is like diagnosing a problem: look at symptoms (the shape of the integrand) and match them to known patterns. The decision tree above is not algorithmic -- two paths may both work, and experience builds intuition about which path is shorter. The only reliable rule is: try the simplest thing first (simplification, then substitution), escalate to more complex techniques only when simpler ones fail.
Connections
Looking back
All major integration techniques (Chapters 4--7) feed into this strategy lesson. It is a synthesis, not a new technique.
Looking ahead
- Integration tables (Section 7.6): A reference tool for integrals that resist the main techniques.
- Numerical methods (Section 7.7): When no closed form exists, numerical integration provides a practical answer.
Back to Techniques of Integration | Next: Using Integration Tables