u-Substitution Explained — Derivation, Examples & Quizzes

Q: How do I choose u?

Pick the 'inside' function — the cluttered part whose derivative (up to a constant factor) also appears in the integrand. In ∫ 2x·e^(x²) dx, u = x² because du = 2x dx is sitting right there.

Q: What if du is off by a constant factor?

That's fine — constants can be adjusted. In ∫ x·e^(x²) dx, du = 2x dx but you only have x dx, so write x dx = du/2 and carry the 1/2 along.

Where it comes from

The problem: the chain rule d/dx [F(g(x))] = F′(g(x))·g′(x) tells you how to differentiate a composition. But integrals like ∫ 2x·e^x² dx stare back at you — no basic rule fits, because the integrand is a composition with its chain-rule factor still attached. u-substitution reads the chain rule backwards.

The tempting “integrate each part”:

Before reading on: differentiate the wrong guess x²·e^x² with the product rule. Do you get 2x·e^x² back — or does an extra term appear that exposes the guess?

\int 2x\cdote x² dx = x 2 \cdot e x² + C ?? the tempting — and wrong — guess

Kill it by differentiating the claimed answer (product rule!): d/dx [x²·e^x²] = 2x·e^x² + x²·2x·e^x² ≠ 2x·e^x². An extra term appears — “integrating each part” is the product-rule fallacy in reverse. The correct move is to notice that 2x is exactly the derivative of the inside function x²:

u = x²

⇒

du = 2x dx

Rename the inner clutter to u. Its differential du = 2x dx is already sitting in the integrand — that is the signal the substitution will work.

∫ 2x·e^x² dx

=

∫ e^u du = e^u + C = e^x² + C

Everything becomes u: the integral collapses to ∫ e^u du, which is trivial. Substitute back at the end.

Intuition: u-substitution is renaming for clarity. The integrand was always “(something) times the derivative of its inside” — writing u just makes that structure visible so the basic rules can fire.

Derivation

Let F′ = f, and set u = g(x). The chain rule applied to F(g(x)) produces exactly the integrand pattern — so integrating reverses it.

d/dx [F(g(x))]

=

F′(g(x))·g′(x) = f(g(x))·g′(x)

Step 1 — chain rule forward. Differentiating the composition F(g(x)) manufactures the pattern: f-of-inside times derivative-of-inside.

∫ f(g(x))·g′(x) dx

=

F(g(x)) + C

Step 2 — read backwards. Since the derivative of F(g(x)) is the integrand, F(g(x)) + C is its antiderivative. This is the substitution formula in disguise.

u = g(x), du = g′(x) dx

⇒

∫ f(u) du = F(u) + C

Step 3 — rename. With u = g(x), the integral ∫ f(g(x))·g′(x) dx is ∫ f(u) du. Evaluating gives F(u) + C = F(g(x)) + C — matching Step 2. ∎

Definite integrals: when x runs from a to b, u runs from g(a) to g(b) — change the bounds when you substitute, then never substitute back. Equivalently, keep x-bounds, substitute back to x, then evaluate. Pick one; do not mix them.

How to use it

Before reading on: after substituting u = x², the integral speaks u but the bounds 0 and 1 still speak x. Before reading: what goes wrong if you evaluate a u-integral at x-bounds — and what are the two legal fixes?

The procedure, every time:

Spot the inside function g(x) — the cluttered part (an exponent, a denominator, inside a root or trig function). Ask: “is its derivative (up to a constant factor) also in the integrand?”
Set u = g(x) and compute du = g′(x) dx.
Rewrite everything in u. Every x must disappear — including the dx. If a stray x remains that is not part of du, the substitution fails; pick a new u.
Fix constant factors: if du = 2x dx but you only have x dx, write x dx = du/2 and carry the 1/2 along. Constants are adjustable; missing variables are fatal.
Integrate in u, then substitute back (indefinite) — or change the bounds to u-values and never look back (definite).

Choosing u: the two best candidates

The inside of a composition: in ∫ x·√(x²+1) dx, u = x²+1 (inside the root). The denominator: in ∫ x/(x²+1) dx, u = x²+1 (du = 2x dx covers the numerator). If neither candidate’s derivative appears, u-substitution is the wrong tool.

Common mistake: substituting back to x and keeping u-bounds in a definite integral — or changing bounds and also substituting back. One road or the other: u-bounds with no back-substitution, or x-bounds with back-substitution. Never both.

Worked examples

Four problems, easiest first. In each one, read every step — the why of each move is the lesson.

Example 1 — the basic move: ∫ 2x·e^x² dx

Choose u. The inside function is x²; its derivative 2x is sitting in the integrand. Set u = x².
Compute du: du = 2x dx — exactly the leftover factor. (Why this u? Because du matches what is there.)
Rewrite: ∫ e^u du = e^u + C.
Substitute back: e^x² + C.
Check: d/dx [e^x²] = 2x·e^x² (chain rule). Matches ✓

Your turn: Compute ∫ 3x²·e^x³ dx.

Answer: e^x³ + C

Inside function x³, derivative 3x² present: set u = x³, du = 3x² dx. Rewrite: ∫ e^u du = e^u + C = e^x³ + C. Check: d/dx [e^x³] = 3x²e^x³. ✓

Example 2 — constant-factor fix: ∫ x·√(x²+1) dx

Choose u. Inside the root: u = x² + 1. Then du = 2x dx.
Fix the constant. We have x dx but need 2x dx: x dx = du/2. (Why legal? Constants factor — only the variable part must match.)
Rewrite: ∫ √u·(du/2) = (1/2)∫ u^1/2 du = (1/2)·(2/3)u^3/2 = (1/3)u^3/2.
Substitute back: (1/3)(x²+1)^3/2 + C.
Check: d/dx = (1/3)(3/2)(x²+1)^1/2·2x = x·√(x²+1). Matches ✓

Common mistake: dropping the 1/2 — writing (2/3)(x²+1)^3/2. The check catches it: differentiating gives 2x·√(x²+1), twice the integrand.

Your turn: Compute ∫ x·(x²+1)² dx.

Answer: (x²+1)³/6 + C

Set u = x² + 1, du = 2x dx, so x dx = du/2. Rewrite: (1/2)∫ u² du = (1/2)(u³/3) = u³/6 = (x²+1)³/6 + C. Check: d/dx = (1/6)·3(x²+1)²·2x = x(x²+1)². ✓

Example 3 — definite, change the bounds: ∫₀¹ x·e^x² dx

Choose u = x², du = 2x dx, so x dx = du/2.
Change the bounds: x = 0 → u = 0; x = 1 → u = 1. (Why? The integral is now in u — x-bounds would be nonsense.)
Rewrite: (1/2)∫₀¹ e^u du = (1/2)[e^u]₀¹ = (e − 1)/2.
Sanity check: on [0,1], x·e^x² runs from 0 to e ≈ 2.718, curving upward; the area should be well under the 1·e rectangle. (e−1)/2 ≈ 0.86 fits. ✓

Common mistake: keeping bounds 0 and 1 but in x after switching to u — here it accidentally works (g(0) = 0, g(1) = 1), which is why the mistake survives until an exam where g(0) ≠ 0. Change bounds every time.

Your turn: Compute ∫₁² x·e^x² dx.

Answer: (e⁴ − e)/2 ≈ 25.94

Set u = x², x dx = du/2. Change bounds: x = 1 → u = 1; x = 2 → u = 4. Rewrite: (1/2)∫₁⁴ e^u du = (1/2)(e⁴ − e). Sanity: on [1,2], x·e^x² runs ≈2.7→≈109 curving upward; 25.94 sits inside. ✓

Example 4 — linear inside: ∫ cos(5x) dx

Choose u = 5x (the inside of the cosine). Then du = 5 dx, so dx = du/5.
Rewrite: ∫ cos u·(du/5) = (1/5)∫ cos u du = (1/5) sin u.
Substitute back: sin(5x)/5 + C.
Check: d/dx = (1/5)·5 cos(5x) = cos(5x). Matches ✓

Common mistake: answering sin(5x) + C — the chain rule’s ×5 needs undoing via ÷5. Whenever the inside is ax + b, expect a ÷a.

Your turn: Compute ∫ sin(3x) dx.

Answer: −cos(3x)/3 + C

Inside 3x: set u = 3x, dx = du/3. Rewrite: (1/3)∫ sin u du = −(1/3) cos u + C = −cos(3x)/3 + C. Check: d/dx [−cos(3x)/3] = −(1/3)(−3 sin(3x)) = sin(3x). ✓

Memorization tips

Say it aloud: “u is the inside; du must be there.” If du (up to a constant) is not in the integrand, abort the substitution.
Constants bend, variables break: a missing 2 is fixable (du/2); a missing x is fatal. That one distinction sorts working substitutions from dead ones.
Bounds travel with the variable: switch to u → switch the bounds. Say “new variable, new bounds” every time.
Linear-inside shortcut: ∫ f(ax+b) dx = (1/a)∫ f(u) du — just divide by a. (It is still u-substitution, with the bookkeeping done once.)
The check is the proof: differentiate your answer with the chain rule. If the chain factor does not reproduce the integrand exactly, recheck du.
Name the failure: ∫ e^x² dx has no x factor — no substitution can conjure one. Recognizing “no elementary antiderivative” is itself a skill.

Final challenge

Five mixed questions — bounds, constant fixes, and the failure case. Score 5/5 and u-substitution is yours.

How to learn a formula here

Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
Work the examples with the answers covered, then uncover one step at a time and compare.
Finish with the final challenge — five mixed questions including the classic traps.
Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.

Frequently asked questions

What is u-substitution?

The chain rule in reverse: if the integrand looks like f(g(x))·g′(x), set u = g(x) so du = g′(x) dx, and the integral becomes the simpler ∫ f(u) du.

How do I choose u?

Pick the “inside” function — the cluttered part whose derivative (up to a constant factor) also appears in the integrand. In ∫ 2x·e^x² dx, u = x² because du = 2x dx is sitting right there.

What if du is off by a constant factor?

That’s fine — constants can be adjusted. In ∫ x·e^x² dx, du = 2x dx but you only have x dx, so write x dx = du/2 and carry the 1/2 along.

Do I change the bounds in a definite integral?

Yes — when you switch to u, the bounds must switch too: x = a becomes u = g(a), x = b becomes u = g(b). Then never substitute back.

When does u-substitution fail?

When the derivative of your candidate u doesn’t appear in the integrand (not even up to a constant). ∫ e^x² dx has no x factor, so no substitution can fix it — it has no elementary antiderivative.

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