Unit 6 · Topic 6.9

6.9 Integrating Using Substitution

u-substitution undoes the chain rule. When an integrand contains an inner function and (up to a constant) that function's derivative, you swap to a new variable u, integrate something simpler, and switch back.

Key terms

u-substitution
chain rule
inner function
du
changing limits of integration

Where it comes from

The chain rule says d/dx F(g(x)) = F′(g(x))·g′(x). Reading that backward: ∫ f(g(x))·g′(x) dx = F(g(x)) + C, where F is an antiderivative of f. Substitution is a tidy way to see this pattern. Let u = g(x), so du = g′(x) dx. The integral becomes ∫ f(u) du, which you can often do with the basic rules.

The steps

Pick u: usually the inside of a power, root, exponent, trig function or denominator.
Find du = u′(x) dx.
Rewrite the whole integral in terms of u and du. If a constant factor is missing, adjust: for example, x dx = ½ du when u = x² + 5.
Integrate with respect to u.
Substitute u = g(x) back in (for indefinite integrals) and add + C.

If an x is left over

If, after substituting, there's still an x left, either your u is a poor choice or you need to solve u = g(x) for x and replace the leftover x too (third example below).

A good test of your choice: after you write du, every piece of the original integrand should be used up exactly once, either inside u or inside du (allowing a constant factor).

Definite integrals: two options

Option 1, change the limits. When you switch to u, also switch the limits: the new lower limit is u(a) and the new upper limit is u(b). Then evaluate in u and never go back to x.

Option 2, go back to x. Find the antiderivative in u, substitute back to x, and then plug in the original limits.

Mixing the two is the classic error: using u-limits with an x-antiderivative, or x-limits with a u-antiderivative. In notation, write the new limits clearly, as in ∫ from u = 5 to u = 9.

Patterns worth recognizing

∫ g′(x)/g(x) dx = ln|g(x)| + C. Example: ∫ 2x/(x² + 1) dx = ln(x² + 1) + C.
Linear inside: ∫ f(ax + b) dx = (1/a)F(ax + b) + C. Example: ∫ cos(3x) dx = (1/3) sin(3x) + C.
∫ tan x dx = ∫ sin x/cos x dx = −ln|cos x| + C (u = cos x).
∫ e^(g(x)) g′(x) dx = e^(g(x)) + C. Example: ∫ x e^(x²) dx = ½ e^(x²) + C.

Worked examples

Try each one yourself first, then open the solution.

Example 1
Indefinite: power of an inner function
Find ∫ x²(x³ + 4)⁵ dx.
Show the solutionHide the solution
1. Step 1: Let u = x³ + 4. Then du = 3x² dx, so x² dx = (1/3) du.
2. Step 2: The integral becomes ∫ u⁵ · (1/3) du = (1/3) · u⁶/6 = u⁶/18.
3. Step 3: Substitute back and add C.
4. Step 4: Check: d/dx [(x³ + 4)⁶/18] = 6(x³ + 4)⁵(3x²)/18 = x²(x³ + 4)⁵.
Answer: (x³ + 4)⁶/18 + C
Example 2
Definite: change the limits
Evaluate ∫₀² x√(x² + 5) dx.
Show the solutionHide the solution
1. Step 1: Let u = x² + 5, so du = 2x dx and x dx = ½ du.
2. Step 2: New limits: when x = 0, u = 5; when x = 2, u = 9.
3. Step 3: The integral becomes ½ ∫ from u = 5 to u = 9 of u^(1/2) du = ½ · (2/3)u^(3/2), evaluated from 5 to 9, which is (1/3)(9^(3/2) − 5^(3/2)).
4. Step 4: 9^(3/2) = 27 and 5^(3/2) = 5√5, so the value is (27 − 5√5)/3 = 9 − (5√5)/3 ≈ 5.273.
Answer: 9 − (5√5)/3 ≈ 5.273
Example 3
Trap: a leftover x
Find ∫ x(2x − 1)⁴ dx.
Show the solutionHide the solution
1. Step 1: Let u = 2x − 1, so du = 2 dx, dx = ½ du.
2. Step 2: The x in front doesn't disappear. Solve for it: x = (u + 1)/2.
3. Step 3: The integral becomes ∫ [(u + 1)/2] · u⁴ · ½ du = ¼ ∫ (u⁵ + u⁴) du = ¼(u⁶/6 + u⁵/5) = u⁶/24 + u⁵/20.
4. Step 4: Substitute back: (2x − 1)⁶/24 + (2x − 1)⁵/20 + C. (You could also expand (2x − 1)⁴ and integrate term by term; both answers differ only by a constant.)
Answer: (2x − 1)⁶/24 + (2x − 1)⁵/20 + C

Common mistakes

Forgetting to convert dx. ∫ cos(3x) dx is not sin(3x) + C; the 1/3 from du = 3 dx matters.
Changing to u but keeping the x-limits, or changing limits and then substituting back to x anyway.
Pulling a variable out like a constant: if u = x² + 1, you can't fix a missing x by dividing by x. Only constant factors can be adjusted.
Writing ln(g(x)) without absolute value when g(x) can be negative, as in ∫ 1/(x − 3) dx = ln|x − 3| + C.

On the exam

Multiple-choice questions often ask which integral in u matches a given integral in x, including the new limits. Check the du factor and both limits.
On free response, writing the substitution (u = …, du = …) and the new limits shows your method and earns credit even if an arithmetic slip follows.

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