Lesson 30: u-Substitution

(a) Opening Question

In Lesson 12 you learned the chain rule: to differentiate a composite function, you multiply the derivative of the outside by the derivative of the inside. For example,

d/dx [(x² + 1)⁴] = 4(x² + 1)³ · 2x = 8x(x² + 1)³

Now run that backward. Suppose someone hands you the integral

∫ 8x(x² + 1)³ dx

You can see the "outside" piece (x² + 1)³ and a factor of 8x. Notice that 2x is exactly the derivative of the inside function x² + 1. That is not a coincidence — it is the fingerprint of a chain-rule derivative.

Your task: Without any new technique, just by reading the chain-rule example above in reverse, write down an antiderivative of 8x(x² + 1)³. Then differentiate your answer to check it. What inside function did you "undo," and how did its derivative show up in the integrand?

(b) Core Concepts

The big idea: u-substitution undoes the chain rule

Every antiderivative rule you know is a derivative rule read backward. Integration by substitution (or u-substitution) is the reverse of the chain rule. When an integrand looks like a function of something, multiplied by the derivative of that something, substitution lets you collapse the whole expression into a single new variable and integrate easily.

The pattern to hunt for is:

∫ f(g(x)) · g'(x) dx

Here g(x) is an inside function and g'(x) is its derivative sitting in the integrand. If we let u = g(x), then du = g'(x) dx, and the integral becomes

∫ f(u) du

which has no composition left — just a plain antiderivative in u.

The method, step by step

1. Choose u. Pick the inside function — usually the quantity inside a power, an exponent, a denominator, or a trig argument — whose derivative also appears in the integrand (possibly off by a constant).
2. Compute du. Differentiate: du = u'(x) dx. Solve for dx or for the grouping you need.
3. Rewrite the integral entirely in terms of u. Every x and every dx must disappear. If a stray x survives, your choice of u was probably wrong.
4. Integrate in u.
5. Back-substitute u = g(x) to return to x (indefinite integrals), and don't forget +C. For definite integrals you instead change the limits (more below).

Worked indefinite example, with the constant-factor adjustment

Find ∫ x(x² + 1)³ dx.

Choose u. The inside function is x² + 1. Let u = x² + 1.

Compute du. Then du = 2x dx. But the integral only has x dx, not 2x dx. No problem — solve for what we have:

du = 2x dx   ⟹   x dx = (1/2) du

This is the constant-factor adjustment: a missing constant can always be supplied, because constants pass freely in and out of integrals. (A missing variable factor cannot — that would mean u was chosen incorrectly.)

Rewrite. Replace (x² + 1)³ with u³ and replace x dx with (1/2) du:

∫ x(x² + 1)³ dx = ∫ u³ · (1/2) du = (1/2) ∫ u³ du

Integrate.

= (1/2) · u⁴/4 + C = u⁴/8 + C

Back-substitute u = x² + 1:

∫ x(x² + 1)³ dx = (x² + 1)⁴ / 8 + C

Verify by differentiating (always do this):

d/dx [(x² + 1)⁴/8] = (1/8) · 4(x² + 1)³ · 2x = x(x² + 1)³ ✓

The derivative returns the original integrand exactly. That is your proof the antiderivative is correct.

The three patterns you will see most

• Power pattern: ∫ (g(x))ⁿ g'(x) dx = (g(x))^{n+1}/(n+1) + C (for n ≠ −1). Choose u = g(x).
• Exponential pattern: ∫ e^{g(x)} g'(x) dx = e^{g(x)} + C. Choose u = g(x) (the exponent).
• Logarithm pattern: ∫ g'(x)/g(x) dx = ln|g(x)| + C. Choose u = g(x) (the denominator); the numerator is its derivative. The absolute value matters: ln|u| is the antiderivative of 1/u for all u ≠ 0.

Trig composites work the same way: in ∫ sin²x cos x dx, the inside is sin x and cos x is its derivative, so u = sin x.

Definite integrals: change the limits (preferred)

For a definite integral ∫ₐᵇ f(g(x)) g'(x) dx you have two valid options.

Method 1 — Change the limits of integration. When you substitute u = g(x), also convert the x-limits into u-limits. The lower limit x = a becomes u = g(a); the upper limit x = b becomes u = g(b). Then you never have to return to x.

Method 2 — Back-substitute to x first. Find the indefinite antiderivative in x (no +C needed for a definite integral), then evaluate at the original x-limits.

Both give the same number. Method 1 is cleaner and is what most AP scorers expect to see.

Worked definite example (both methods)

Evaluate ∫₀² x(x² + 1)³ dx.

Use u = x² + 1, so x dx = (1/2) du.

Method 1 — change limits. When x = 0, u = 0² + 1 = 1. When x = 2, u = 2² + 1 = 5. The integral becomes

∫₀² x(x² + 1)³ dx = (1/2) ∫₁⁵ u³ du = (1/2) · [u⁴/4]₁⁵
                  = (1/8)(5⁴ − 1⁴) = (1/8)(625 − 1) = 624/8 = 78

Method 2 — back-substitute. The antiderivative in x is (x² + 1)⁴/8, so

[(x² + 1)⁴/8]₀² = (5⁴)/8 − (1⁴)/8 = 625/8 − 1/8 = 624/8 = 78

Same answer, 78. Notice the #1 limit-change trap: if you change the limits to u-values, you must not also plug the original x-limits into a u-expression. Pick one method and stay consistent.

(c) Worked Examples

Example 1 — Basic power pattern (foundation) [NO CALC]

Problem. Find ∫ 3x²(x³ + 5)⁴ dx.

Strategy. The inside function is x³ + 5; its derivative is 3x², which appears exactly. Power pattern.

Solution. Let u = x³ + 5, so du = 3x² dx. The integral becomes

∫ u⁴ du = u⁵/5 + C = (x³ + 5)⁵/5 + C

Verify. d/dx [(x³ + 5)⁵/5] = (1/5)·5(x³ + 5)⁴·3x² = 3x²(x³ + 5)⁴ ✓

Because 3x² was already present, no constant adjustment was needed.

Example 2 — Exponential composite (procedural) [NO CALC]

Problem. Find ∫ x e^{x²} dx.

Strategy. The exponent x² is the inside function; its derivative 2x is present up to the constant 2.

Solution. Let u = x², so du = 2x dx, giving x dx = (1/2) du.

∫ x e^{x²} dx = (1/2) ∫ e^u du = (1/2) e^u + C = (1/2) e^{x²} + C

Verify. d/dx [(1/2)e^{x²}] = (1/2)e^{x²}·2x = x e^{x²} ✓

Example 3 — The g'/g → ln pattern (procedural) [NO CALC]

Problem. Find ∫ 2x/(x² + 1) dx.

Strategy. The denominator x² + 1 has derivative 2x, which is exactly the numerator. This is the log pattern.

Solution. Let u = x² + 1, so du = 2x dx.

∫ 2x/(x² + 1) dx = ∫ (1/u) du = ln|u| + C = ln(x² + 1) + C

(We may drop the absolute value here because x² + 1 > 0 for all x, but write ln|u| as a habit.)

Verify. d/dx [ln(x² + 1)] = 1/(x² + 1) · 2x = 2x/(x² + 1) ✓

Example 4 — Definite integral with changed limits (AP level) [NO CALC]

Problem. Evaluate ∫₀^{π/2} sin²x cos x dx.

Strategy. The inside function is sin x, and cos x (its derivative) is present. Change the limits.

Solution. Let u = sin x, so du = cos x dx. New limits: when x = 0, u = sin 0 = 0; when x = π/2, u = sin(π/2) = 1.

∫₀^{π/2} sin²x cos x dx = ∫₀¹ u² du = [u³/3]₀¹ = 1/3 − 0 = 1/3

Justification of the limit change. Because u = sin x is continuous and we substituted on a closed interval, evaluating the u-integral from u = sin 0 = 0 to u = sin(π/2) = 1 is equivalent to evaluating the original integral from x = 0 to x = π/2. The answer is 1/3.

(d) Common Mistakes

Forgetting to convert dx via du. Students replace g(x) with u but leave a leftover dx, then "integrate" a mishmash. Why it's wrong: an integral in u must contain du, not dx. Fix: always write du = g'(x) dx explicitly and substitute it; if any dx or x survives, stop and recheck your u.

Dropping the constant factor. From du = 2x dx, students treat x dx as if it equals du. Why it's wrong: x dx = (1/2) du, not du — you lose a factor of 1/2. Fix: solve du for the exact grouping in the integrand and carry the constant out front.

On a definite integral, changing the limits AND back-substituting. Students convert the limits to u-values but then also plug the original x-limits into a u-expression (or vice versa). Why it's wrong: you end up evaluating at the wrong points. Fix: pick ONE method. If you change limits, evaluate the u-antiderivative at the u-limits and never return to x. If you back-substitute, use the original x-limits.

Forgetting +C on an indefinite integral. Why it's wrong: ∫ f dx is a family of antiderivatives; omitting +C loses the answer point on the AP exam. Fix: the instant you finish integrating an indefinite integral, write +C.

Choosing a u whose derivative isn't present. Students pick u = x² for ∫ (x² + 1)³ dx (note: no x factor). Why it's wrong: du = 2x dx, but there's no x to absorb, leaving a stray x. Fix: u must be an inside function whose derivative appears up to a constant — if it doesn't, substitution won't work directly.

(e) Practice Problems

Solutions in the Answer Key. Justification questions are marked ★.

Find ∫ (ln x)/x dx.

Evaluate ∫₀² x(x² + 1)³ dx using the change-of-limits method. Show the new limits.

★A student computes ∫₀² 2x(x²) dx by letting u = x² and writing ∫₀² u du = [u²/2]₀² = 2. Identify the error in the limits and give the correct value with justification.

★Explain why ∫ (x² + 1)³ dx cannot be evaluated by the substitution u = x² + 1 directly, but ∫ x(x² + 1)³ dx can. Reference what du requires.

Find ∫ x/√(x² + 16) dx (indefinite).

(f) AP Exam Focus

> Free Response Question (No calculator) — 9 points total

Water flows into a holding tank. For 0 ≤ t ≤ 4 hours, the rate at which water enters the tank is modeled by

R(t) = 6t/(t² + 1)   liters per hour.

The tank contains 10 liters of water at time t = 0.

(a) (2 points) Find the rate at which water is entering the tank at time t = 4. Indicate units.

(b) (2 points) Find the time t in 0 < t < 4 at which R(t) attains its maximum value. Justify your answer.

(c) (3 points) Find the total amount of water, in liters, that enters the tank during the interval 0 ≤ t ≤ 4. Show the setup and use a substitution; give an exact value.

(d) (2 points) Find the total amount of water in the tank at time t = 4. Using the result of part (c), determine whether the average rate of flow over [0, 4] exceeds R(4), and interpret what that comparison means.

Model Solution

(a) Evaluate the model directly:

R(4) = 6(4)/(4² + 1) = 24/17 ≈ 1.412 liters per hour.

Water is entering at a rate of 24/17 ≈ 1.412 liters per hour at t = 4. (2 pts: 1 for value, 1 for units.)

(b) Differentiate using the quotient rule:

R'(t) = [6(t² + 1) − 6t(2t)] / (t² + 1)²
      = [6t² + 6 − 12t²] / (t² + 1)²
      = 6(1 − t²) / (t² + 1)².

R'(t) = 0 requires 1 − t² = 0, so t = 1 (rejecting t = −1, outside the domain). On (0, 1), 1 − t² > 0 so R' > 0; on (1, 4), 1 − t² < 0 so R' < 0. Since R' changes from positive to negative at t = 1, R has a relative maximum there, and because it is the only critical point in (0, 4), the maximum on the interval occurs at t = 1. (2 pts: 1 for t = 1, 1 for sign-change justification.)

(c) The total water entering over [0, 4] is the accumulation

∫₀⁴ R(t) dt = ∫₀⁴ 6t/(t² + 1) dt.

Let u = t² + 1, so du = 2t dt and 6t dt = 3 du. Change limits: t = 0 ⟹ u = 1; t = 4 ⟹ u = 17.

∫₀⁴ 6t/(t² + 1) dt = 3 ∫₁^{17} (1/u) du = 3[ln|u|]₁^{17}
                   = 3(ln 17 − ln 1) = 3 ln 17 liters.

So 3 ln 17 ≈ 8.500 liters enter the tank. (3 pts: 1 for the integral setup, 1 for a correct substitution/antiderivative, 1 for the exact value 3 ln 17.)

(d) The tank starts with 10 liters and gains 3 ln 17 liters, so at t = 4 it holds

10 + 3 ln 17 ≈ 18.500 liters.

The average rate of flow over [0, 4] is

(1/(4 − 0)) ∫₀⁴ R(t) dt = (3 ln 17)/4 ≈ 2.125 liters per hour.

Since 2.125 > 24/17 ≈ 1.412, the average rate of flow over the interval exceeds the instantaneous rate at t = 4. This means the flow was, on average, faster than its end-of-interval value — consistent with the rate peaking at t = 1 and then declining toward t = 4. (2 pts: 1 for 10 + 3 ln 17, 1 for the average-vs-R(4) comparison with interpretation.)

Scoring Commentary — where students lose points

• Part (a): Omitting units is the most common single-point loss on rate questions. Always label "liters per hour."
• Part (b): Writing only "t = 1" with no sign analysis earns the value point but loses the justification point. The phrase that scores is "R' changes from positive to negative," not "it's a max."
• Part (c): The classic error is not changing the limits after substituting u = t² + 1, then evaluating 3[ln u] at 0 and 4 (the t-limits) instead of 1 and 17 — giving 3 ln 17 accidentally only if lucky, otherwise nonsense. Equally common: dropping the factor that makes 6t dt = 3 du.
• Part (d): Students forget to add the initial 10 liters, reporting 3 ln 17 as the tank total instead of the amount added. Read carefully: net change vs. total.