AP Calculus AB and BC

U-Substitution Helper: Pick u, See Why It Works

Pick u to be the inside function whose derivative already appears in the integrand as a factor. Test a candidate by differentiating it: if du (up to a constant) is present, the choice works and the integral collapses; otherwise pick a different u. This tool runs that test and hints instead of solving.

Problem 1 of 8

\int 2x\left(x^2+1\right)^4\,dx

Which substitution makes this integral simpler?

This helper drills the single decision that makes or breaks a substitution: which part of the integrand to call $u$ . You pick a candidate from the pieces on screen, and the tool differentiates it, checks whether $du$ is available in the integrand, and rewrites the integral in $u$ when the choice is sound. When the choice is poor it stops and returns a hint instead of an answer, so the goal is to learn the reasoning, not to be handed a solved integral. Substitution is CED Topic 6.9 (FUN-6.D.1), and it shows up on both the AB and BC exams, usually as one step buried inside a larger accumulation or area problem.

\int f(g(x))\,g'(x)\,dx = \int f(u)\,du, \qquad u = g(x)

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 a composite piece next to the derivative of its inner function is built to be un-chained. Every $u$ you would ever pick comes from the same short list of spots, all of them the inside of a composite: the base of a power like $(x^2+3)^4$ , the expression under a root, the exponent on $e$ , the argument of a $\sin$ , $\cos$ , or $\tan$ , or a denominator. Each inner function is a candidate. The tool surfaces those candidates; your job is to decide which one has its derivative sitting elsewhere in the integrand.

The test the tool runs

Differentiate your candidate. If the result, times any constant, is already a factor in the integrand, then $du$ can be formed and the substitution works. If differentiating produces a factor that appears nowhere, that candidate fails and you get a hint.

When your pick passes, read what the rewrite shows: every $x$ disappears, including inside the $dx$ , and the integral collapses into a basic form in $u$ . A leftover constant is harmless, because you fix it by multiplying and dividing. If $du = 2x\,dx$ and the integral only has $x\,dx$ , then $x\,dx = \frac{1}{2}\,du$ . A leftover variable is not harmless: if an $x$ still survives after substituting, the rewrite is incomplete and the choice usually has to change.

Two failure signals trigger a hint. The derivative is missing: set $u = x^2$ in $\int e^{x^2}\,dx$ and $du = 2x\,dx$ needs an $x$ that is not there, so the substitution stalls (this integrand has no elementary antiderivative at all). A stray variable is left behind: the rewrite still carries an $x$ that $du$ cannot absorb. Reading which signal fired tells you what to do next, either try a different candidate or, in the stray-variable case, solve $u = g(x)$ for $x$ and substitute that too.

Worked example

Testing candidates: why u equals x^4 beats u equals x^3

Evaluate $\int x^3 \cos(x^4)\,dx$ .

List the candidates the way the tool does. The integrand has two pieces, the factor $x^3$ and the composite $\cos(x^4)$ . The plausible candidates for $u$ are $u = x^3$ (the standalone factor) and $u = x^4$ (the inner function of the composite). Test each by differentiating and checking whether its $du$ is present.
Test $u = x^3$ . Then $\frac{du}{dx} = 3x^2$ , so $du = 3x^2\,dx$ . The factor $x^2$ is available, since $x^3 = x \cdot x^2$ , but pulling out $x^2\,dx$ leaves a stray $x$ : $x^3\,dx = \frac{x}{3}\,du$ . That leftover $x$ cannot be written in terms of $u = x^3$ (the composite's argument $x^4$ is not a function of $x^3$ either), so the rewrite still carries an $x$ . This is the stray-variable signal, and the tool returns a hint rather than a clean rewrite.
Test $u = x^4$ . Then $\frac{du}{dx} = 4x^3$ , so $du = 4x^3\,dx$ . The factor $x^3\,dx$ is exactly what sits next to the cosine, matched up to the constant $4$ . Solve for that piece: $x^3\,dx = \frac{1}{4}\,du$ . This candidate passes.
Rewrite the integral entirely in $u$ : $\int x^3 \cos(x^4)\,dx = \int \cos(x^4)\cdot x^3\,dx = \int \cos u \cdot \frac{1}{4}\,du$ . No $x$ remains anywhere, which confirms the substitution is complete.
Factor the constant out front: $\int \cos u \cdot \frac{1}{4}\,du = \frac{1}{4}\int \cos u\,du$ .
Integrate in $u$ : $\frac{1}{4}\int \cos u\,du = \frac{1}{4}\sin u$ .
Back-substitute $u = x^4$ and add the constant: $\frac{1}{4}\sin(x^4) + C$ .
Check by differentiating: $\frac{d}{dx}\left[\frac{1}{4}\sin(x^4)\right] = \frac{1}{4}\cos(x^4)\cdot 4x^3 = x^3\cos(x^4)$ , which is the original integrand.

$\dfrac{1}{4}\sin(x^4) + C$

Frequently asked questions

How do you choose u in u-substitution?

Pick the inner function of a composite whose derivative also appears in the integrand. Test a candidate by differentiating it: if the result, times any constant, is already a factor in the integral, $du$ can be formed and the choice works. The usual candidates are bases of powers, radicands, exponents of $e$ , trig arguments, and denominators.

Why did the helper flag my u with a hint instead of solving?

Because your candidate did not pass the test. Either its derivative is missing (differentiating produced a factor that is nowhere in the integrand, so $du$ cannot be built), or the rewrite left a stray $x$ that $du$ cannot absorb. The hint names which signal fired so you know whether to try another candidate or to solve $u = g(x)$ for $x$ .

Does it matter which candidate I try first?

No. The differentiate-and-check test is quick, and only the candidate whose derivative is present produces a clean rewrite with no $x$ left. Try any candidate; the tool will pass the good one and hint on the rest. In the stray-variable case you may also need to solve $u = g(x)$ for $x$ before every $x$ clears.