3.5 Other Strategies for Integration

Big idea

Up to now every problem arrived labelled. A section on parts handed you parts problems; a section on partial fractions handed you rational functions. On an exam, and in every application afterwards, integrals arrive unlabelled, and choosing the method is most of the work. This section is about that choice.

Differentiation is an algorithm. Every elementary function has an elementary derivative, and a finite list of rules computes it without any decisions. Integration is a search. There is no rule that says which technique applies, the same integral often yields to two different routes, and some integrands have no elementary antiderivative at all. Recognizing structure is therefore a skill you build by classifying, not by memorizing.

The good news is that the space of possibilities is small. Simplify algebraically, try a substitution, look for a product that wants parts, check whether the integrand is rational, check whether it involves a quadratic under a root. Five questions, asked in that order, dispose of nearly everything in a first-year course. When one of them succeeds you are done deciding; when one fails you have lost thirty seconds.

The last idea in the section is a boundary rather than a technique. A theorem of differential algebra guarantees that certain perfectly respectable functions have no antiderivative expressible with the elementary operations. This is not a gap in your training; it is a fact about the functions. When you meet one, the correct move is to switch tools: a numerical method, a series expansion, or a named special function.

Decoder

Choosing a method is a matter of reading the structure of the integrand: what it is built from, which piece is the derivative of another, and whether any algebraic rewriting makes a known form appear.

The phrase to hold onto is derivative of another. That single relationship is the signal for substitution, which is the first technique to try because it is the cheapest and because it often converts a hard-looking integral into an easy one. Scan the integrand for a composite function and ask whether the derivative of the inner function is sitting outside, up to a constant.

If no such pairing exists, the second question is whether the integrand is a product of two unrelated types, which is the signal for parts, and the third is whether it is a quotient of polynomials, which is the signal for partial fractions. These signals are structural, not numerical: they depend on what kind of functions appear, not on the specific coefficients.

The classic mistake is committing too early and grinding. If two lines of work have made the integrand more complicated rather than less, the method was wrong and continuing will not fix it. Stop, return to the original integrand, and ask the next question on the list.

Definitions and results

Simplify first. Before choosing a technique, expand products, split fractions with a single term in the denominator, cancel common factors, and apply trigonometric identities. A surprising number of integrals stop being hard at this stage.

Substitution. Look for $f(g(x))g'(x)$ up to a constant multiple. Powers of a linear expression, exponentials with a nonlinear exponent, and anything where a denominator’s derivative sits in the numerator all fall here. This is the first technique to try.

Integration by parts. Look for a product of two different function types, one of which improves under differentiation. Polynomial times exponential, polynomial times trigonometric, and anything containing a logarithm or an inverse trigonometric function are the standard cases.

Partial fractions. Look for a quotient of polynomials. Divide first if the numerator’s degree is at least the denominator’s, then factor the denominator and decompose.

Trigonometric substitution. Look for $\sqrt{a^2-x^2}$, $\sqrt{a^2+x^2}$, or $\sqrt{x^2-a^2}$, including after completing the square. If the derivative of the quadratic already appears as a factor, use a plain substitution instead.

Combining methods. Many integrals need two techniques in sequence. A substitution can reduce an integral to a parts problem, and a parts step can produce a rational function. Treat the result of each step as a fresh integral and ask the questions again.

More than one route. Different choices can lead to answers that look different but differ only by a constant. For instance $\int \sin x\cos x\,dx$ gives $\frac{1}{2}\sin^2 x + C$ with one substitution and $-\frac{1}{2}\cos^2 x + C$ with another, and the two differ by $\frac12$. Both are correct.

Nonelementary antiderivatives. Some continuous functions have no antiderivative expressible in terms of powers, roots, exponentials, logarithms and trigonometric functions. Standard cases include

$$ \int e^{-x^2}\,dx, \qquad \int \frac{\sin x}{x}\,dx, \qquad \int \frac{dx}{\ln x}, \qquad \int \sqrt{1+x^3}\,dx $$

These are genuinely impossible, not merely difficult. Definite versions are computed numerically.

Worked examples

Two integrals that differ by one factor

Compare $\int \frac{x}{x^2+9}\,dx$ with $\int \frac{dx}{x^2+9}$.

The first has the derivative of the denominator, up to the factor $2$, sitting in the numerator. That is the substitution signal. With $u = x^2+9$ and $du = 2x\,dx$,

$$ \int \frac{x}{x^2+9}\,dx = \frac{1}{2}\ln\big(x^2+9\big) + C $$

The second has no such factor, so substitution has nothing to cancel. It is instead a standard form with $a = 3$:

$$ \int \frac{dx}{x^2+9} = \frac{1}{3}\arctan\frac{x}{3} + C $$

One factor of $x$ separates a logarithm from an arctangent. Read the numerator before you read anything else.

A substitution that sets up parts

Evaluate $\int x^3 e^{x^2}\,dx$.

Parts applied directly fails: $e^{x^2}$ cannot be antidifferentiated, so it cannot be $dv$, and taking $u = e^{x^2}$ raises the power of $x$. But the exponent suggests $u = x^2$, with $du = 2x\,dx$, and there is an $x$ available to pair with $dx$. Writing $x^3 = x^2\cdot x$,

$$ \int x^3 e^{x^2}\,dx = \frac{1}{2}\int u e^{u}\,du $$

Now it is an ordinary parts problem, giving $\frac{1}{2}\big(u e^u - e^u\big)$. Substituting back,

$$ \int x^3 e^{x^2}\,dx = \frac{1}{2}e^{x^2}\big(x^2 - 1\big) + C $$

Check by differentiating: $\frac{1}{2}\left[2x e^{x^2}(x^2-1) + e^{x^2}(2x)\right] = \frac{1}{2}e^{x^2}\left[2x^3 - 2x + 2x\right] = x^3e^{x^2}$.

Algebra before calculus

Evaluate $\int \frac{dx}{1+\sin x}$.

None of the four techniques applies to this as written. The rewriting move for a sum in a denominator is to multiply by the conjugate:

$$ \frac{1}{1+\sin x}\cdot\frac{1-\sin x}{1-\sin x} = \frac{1-\sin x}{1-\sin^2 x} = \frac{1-\sin x}{\cos^2 x} $$

Splitting the fraction gives $\sec^2 x - \sec x\tan x$, and both terms are standard derivatives read backwards:

$$ \int \frac{dx}{1+\sin x} = \tan x - \sec x + C $$

Differentiating returns $\sec^2 x - \sec x\tan x$, which recombines to the original integrand.

When no elementary answer exists

Compare $\int 2x e^{x^2}\,dx$ with $\int e^{x^2}\,dx$.

The first is a one-line substitution with $u = x^2$, giving $e^{x^2} + C$. The second differs only by the missing factor $2x$, and it has no elementary antiderivative whatsoever. No amount of cleverness will produce one, because the impossibility is a theorem.

What you do instead depends on the question. For a definite integral such as $\int_0^1 e^{x^2}\,dx$, use a numerical rule. For an approximation near a point, expand the integrand as a power series and integrate term by term. For work in probability, the antiderivative of $e^{-x^2}$ is given a name, the error function, and tabulated. Recognizing that an integral is in this family is itself the correct answer, and it saves you an hour of searching for a technique that does not exist.

Practice

Start with substitution, the first question on the list and the most common answer to it. These problems reward scanning for an inner function whose derivative is already present.

Next, products that want parts. Decide which factor improves under differentiation before you write anything, and watch for problems that need a substitution first.

Last, rational integrands. Confirm properness, factor the denominator, and choose the right template; some of these collapse to a substitution once you look at the numerator.

Quiz

Six items on identifying the right method, combining two techniques, rewriting algebraically, and recognizing integrals with no elementary antiderivative.