Difficulty: Intermediate | Prerequisites: Section 13.1 (indefinite integrals), chain rule from differentiation.
Substitution is the integration counterpart of the chain rule. When an integrand is a composite function, straightforward power-rule or exponential integration will not work directly. Substitution lets you simplify a complicated integral by replacing a chunk of the integrand with a single variable u, integrating in terms of u, then converting back. This is the most-used integration technique in the course and appears on nearly every exam.
Pick the "inner function" as u, compute du, rewrite the integral entirely in terms of u and du, integrate, then substitute back. Shortcut formulas exist for simple linear substitutions like e^(ax) and (ax + b)^n.
u-substitution
A technique for evaluating integrals by letting u equal an "inner function" inside the integrand, then rewriting the entire integral in terms of u and du. Think of it as undoing the chain rule: the chain rule produced the composite structure, and substitution peels it apart.
Inner function
The part of the integrand that sits inside another operation (inside an exponent, under a radical, raised to a power, inside a logarithm). This is what you set equal to u. In simple terms, it is the piece whose derivative also appears as a factor in the integrand.
du
The differential of u, found by differentiating u with respect to x and writing du = u′(x) dx. You need to match or adjust for this factor in the integrand to complete the substitution.
Shortcut formula (linear substitution)
Pre-built results for integrals where the inner function is linear (ax + b). These save time on exams: ∫ e^(ax) dx = (1/a)e^(ax) + C, ∫ (ax + b)^n dx = (ax + b)^(n+1) / [a(n + 1)] + C, and ∫ 1/(ax + b) dx = (1/a) ln|ax + b| + C.
Identify the inner function and let u = that expression
Differentiate: du = u′(x) dx
Solve for dx (or match the remaining x-terms with du)
Rewrite the entire integral in terms of u and du, with no x remaining
Integrate with respect to u
Substitute back, replacing u with the original expression in x
Add + C
Look for a function whose derivative (or a constant multiple of it) also appears in the integrand
Common patterns:
e^(something): let u = the exponent
(something)^n: let u = the base
ln(x)/x: let u = ln x, so du = 1/x dx
√(something): let u = the expression under the radical
When the inner function is of the form ax + b (linear), you can skip the full substitution and use these directly:
∫ e^(ax) dx = (1/a) e^(ax) + C
∫ (ax + b)^n dx = (ax + b)^(n+1) / [a(n + 1)] + C, for n ≠ −1
∫ 1/(ax + b) dx = (1/a) ln|ax + b| + C
These are labelled "can also use short-cut formula" on the exam review for problems (a), (b), and (c) in Section 13.2.
Sometimes you need to adjust for a constant factor. For example, in ∫ 4x e^(x² − 3) dx, letting u = x² − 3 gives du = 2x dx. The integrand has 4x dx = 2 · (2x dx) = 2 du. So the integral becomes 2 ∫ eᵘ du = 2eᵘ + C = 2e^(x² − 3) + C.
Let u = x² + 1, so du = 2x dx, which means x dx = du/2.
The integral becomes (1/2) ∫ u^10 du = (1/2) · u^11/11 + C = (1/22)(x² + 1)^11 + C.
Let u = ln x, so du = (1/x) dx.
The integral becomes ∫ 3u du = (3/2)u² + C = (3/2)(ln x)² + C.
\int e^{ax}\,dx = \frac{1}{a}\,e^{ax} + C\int (ax+b)^n\,dx = \frac{(ax+b)^{n+1}}{a(n+1)} + C \quad (n \neq -1)\int \frac{1}{ax+b}\,dx = \frac{1}{a}\ln|ax+b| + CGeneral substitution pattern:
\int f(g(x))\,g'(x)\,dx = \int f(u)\,du \quad \text{where } u = g(x)Substitution is used whenever a rate-of-change model involves composite functions, which is most real models. In pharmacokinetics, drug concentration decays as e^(−kt), and integrating that expression (a substitution with u = −kt) gives total drug exposure over time.
Students forget to convert dx to du. If you still have dx in your integral after substitution, the substitution is incomplete.
Students pick the wrong u. The key test: does du (or a constant multiple of it) appear as a factor in the remaining integrand? If not, try a different choice.
Students forget to substitute back at the end. Your final answer must be in terms of x, not u.
Students try substitution when simple algebra (expanding, splitting) would be faster. Not every integral needs substitution.
⚠️ Know the three shortcut formulas for linear substitutions cold. They save time and reduce errors.
⚠️ Show every step: define u, compute du, rewrite, integrate, substitute back. Partial credit requires visible work.
⚠️ Problems where the derivative of u appears with a different constant multiplier (e.g. 4x dx when du = 2x dx) are a favourite test of whether you can adjust the constant correctly.
True or false: for ∫ 16e^(−3x) dx, you can use the shortcut formula directly. (True. The answer is −(16/3)e^(−3x) + C.)
Fill in the blank: if u = x² − 3, then du = ____. (2x dx)
True or false: ∫ (3x² + 1)(x³ + x − 2)⁹ dx requires u = 3x² + 1. (False. Let u = x³ + x − 2, because its derivative is 3x² + 1.)
Fill in the blank: after integrating ∫ eᵘ du, you must ____ before writing the final answer. (Substitute back for u in terms of x)
True or false: ∫ 1/(2x − 5) dx = ln|2x − 5| + C. (False. You need the 1/2 factor: (1/2) ln|2x − 5| + C.)
Q: Find ∫ 16e^(−3x) dx.
A: Shortcut formula with a = −3: (16/−3)e^(−3x) + C = −(16/3)e^(−3x) + C.
Q: Find ∫ (5x − 2)³ dx.
A: Shortcut with a = 5, n = 3: (5x − 2)⁴ / (5 × 4) + C = (1/20)(5x − 2)⁴ + C.
Q: Find ∫ 1/(2x − 5) dx.
A: Shortcut with a = 2: (1/2) ln|2x − 5| + C.
Q: Find ∫ 4x e^(x² − 3) dx.
A: Let u = x² − 3, du = 2x dx, so 4x dx = 2 du. The integral becomes 2 ∫ eᵘ du = 2eᵘ + C = 2e^(x² − 3) + C.
Q: Find ∫ x(x² + 1)^10 dx.
A: Let u = x² + 1, du = 2x dx, so x dx = du/2. The integral becomes (1/2) ∫ u^10 du = (1/2)(u^11/11) + C = (1/22)(x² + 1)^11 + C.
Q: Find ∫ 15x√(−x² + 7) dx.
A: Let u = −x² + 7, du = −2x dx, so 15x dx = −(15/2) du. The integral becomes −(15/2) ∫ u^(1/2) du = −(15/2)(2/3)u^(3/2) + C = −5(−x² + 7)^(3/2) + C.
Q: Find ∫ (3x² + 1)(x³ + x − 2)⁹ dx.
A: Let u = x³ + x − 2, du = (3x² + 1) dx. The integral becomes ∫ u⁹ du = u^10/10 + C = (1/10)(x³ + x − 2)^10 + C.
Q: Find ∫ (3 ln x)/x dx.
A: Let u = ln x, du = (1/x) dx. The integral becomes ∫ 3u du = (3/2)u² + C = (3/2)(ln x)² + C.
Substitution builds on the chain rule from differentiation (Chapters 11 and 12). It is also essential for evaluating definite integrals in Section 13.3, where you either change the limits of integration to match u or substitute back before plugging in bounds. Area-between-curves problems in Section 14.2 sometimes require substitution to evaluate the resulting integral.
u-substitution, integration by substitution, chain rule reverse, inner function, shortcut formula, linear substitution, composite function integral, MAT 210, brief calculus, Section 13.2, ASU calculus, change of variable