U-Substitution: How to Choose u and Integrate It

When u-substitution is the right tool

Reach for u-substitution (CED Topic 6.9) when the integrand is a composite function multiplied by the derivative of its inside. Substitution is the chain rule run backward: the chain rule turns $F(g(x))$ into $F'(g(x)) \cdot g'(x)$, so an integral that carries both a composite piece and the derivative of its inner function is set up to be un-chained.

The recognition signal is a two-part pattern. First, find an inner function $g(x)$ tucked inside something else: the base of a power, the radicand of a root, the exponent on $e$, the argument of a trig function, or a denominator. Second, check whether $g'(x)$, the derivative of that inner function, also shows up in the integrand. It only has to match up to a constant multiple, since constants are easy to carry through. When both parts are present, substitution collapses the whole integral into a basic one. On the AP exam this technique appears on both multiple-choice and free-response questions, often buried inside a larger accumulation or area problem.

Choosing u: pick the inside

The choice of $u$ decides everything, and the rule is short: let $u$ be the inner function whose derivative appears. In $\int x(x^2+3)^4\,dx$ the inside is $x^2+3$ because it is raised to a power, and its derivative $2x$ is sitting right there as a factor. In $\int \frac{\ln x}{x}\,dx$ the inside is $\ln x$, since its derivative $\frac{1}{x}$ is the other factor. In $\int \cos(5x)\,dx$ the inside is $5x$, the argument of the cosine.

Work from the outside in. Ask what function is wrapped around another, then name the wrapped part $u$. A reliable checklist of candidates: the base of a power such as $(x^2+3)^4$; the expression under a root; the exponent of $e^{(\cdots)}$; the argument of $\sin$, $\cos$, or $\tan$; and the denominator of a fraction. After you pick, differentiate to confirm $g'(x)$ is available. A leftover constant is harmless, since you fix it by multiplying and dividing. A leftover variable is a warning sign that the choice may be wrong.

One subtlety: the inside is not always the most complicated-looking chunk. It is specifically the function whose derivative you can find elsewhere. In $\int x^3 \cos(x^4)\,dx$ the tempting choice is $u = x^3$, but its derivative $3x^2$ is not in the integrand; the right choice is $u = x^4$, whose derivative $4x^3$ matches the $x^3$ factor. Let the available derivative, not the visual bulk, decide.

The five steps

Once $u$ is chosen, the procedure is mechanical.

1. Set $u = g(x)$ and differentiate to get $du = g'(x)\,dx$.
2. Solve for the exact piece the integral contains. If $du = 2x\,dx$ but the integral has $x\,dx$, then $x\,dx = \frac{1}{2}\,du$.
3. Rewrite the integral so that no $x$ remains anywhere, including inside the $dx$. A surviving $x$ means the substitution is incomplete.
4. Integrate in terms of $u$ using the basic antiderivative rules.
5. For an indefinite integral, back-substitute $u = g(x)$ and add $+C$. For a definite integral, handle the bounds instead.

Steps two and three are where substitution succeeds or fails. If you can express every $x$-piece in terms of $u$ and $du$, the integral was a genuine substitution. If you cannot, the setup is telling you to reconsider $u$.

Definite integrals: convert the bounds

For a definite integral, the essential-knowledge point of Topic 6.9 (FUN-6.D.2) is that substitution requires corresponding changes to the limits of integration. The bounds written on the integral are $x$-values. The moment you switch the variable to $u$, those numbers no longer describe where you are, so you must deal with them on purpose.

There are two safe routes. The cleaner one: convert the limits by plugging each original bound into $u = g(x)$, then evaluate the antiderivative entirely in $u$ and never return to $x$. The alternative: find the antiderivative, back-substitute to $x$, then use the original $x$-bounds. Pick one and commit to it.

What a failed choice looks like

A wrong $u$ announces itself quickly, which is useful, because you can bail out before wasting time. Three failure patterns cover almost every case.

The inner derivative is missing. Set $u = x^2$ in $\int e^{x^2}\,dx$ and you get $du = 2x\,dx$, but there is no $x$ anywhere in the integrand to absorb it. The substitution stalls, and this particular integral has no elementary antiderivative at all. The lesson is that the derivative of your $u$ has to actually be present, not just wished for.

A variable factor is left behind. If rewriting the integral in $u$ still leaves a stray $x$, the choice usually needs rescuing. Sometimes you can solve $u = g(x)$ for $x$ and substitute that expression too, as in Example 3. If that does not clear the $x$, abandon the choice and try a different $u$ or a different technique from Topic 6.14.

The constant gets dropped. When $du = 2x\,dx$ but the integrand carries only $x\,dx$, the factor of $\frac{1}{2}$ is easy to forget. Carry it through, and always confirm by differentiating your answer, which should reproduce the original integrand exactly.