U-substitution

U-substitution reverses the chain rule. If the integrand contains a function g(x) and its derivative g′(x), let u = g(x), rewrite the integral in u, integrate, and substitute back.

By the Calcist team · Updated October 26, 2026 · About a 6-minute read

The rule

Let u = g(x), where g′(x) is continuous on an interval. Then

∫ f(g(x)) g′(x) dx = ∫ f(u) du = F(u) + C = F(g(x)) + C

where F is an antiderivative of f. In practice: replace g(x) with u and g′(x) dx with du.

How to choose u

Look for an expression g(x) inside the integrand whose derivative g′(x) also appears (up to a constant factor). Common choices: the inside of a power, a root, an exponent or a trig function.
Set u = g(x) and compute du = g′(x) dx.
Rewrite the whole integral in terms of u. No x should be left.
If the new integral isn't easier, try a different u.
Integrate, then write the answer in terms of x.

Example 1: ∫ 6x(3x² + 4)⁴ dx

The inside of the power is 3x² + 4, and its derivative 6x is right there. Let u = 3x² + 4, so du = 6x dx.
The integral becomes ∫ u⁴ du.
Integrate: u⁵/5 + C.
Substitute back: (3x² + 4)⁵/5 + C.

Example 2: ∫ z√(z² − 5) dz (adjusting a constant)

Let u = z² − 5, so du = 2z dz. The integrand has z dz, not 2z dz, so z dz = (1/2) du.
The integral becomes (1/2) ∫ u^1/2 du.
Integrate: (1/2) · (2/3) u^3/2 = (1/3) u^3/2 + C.
Answer: (1/3)(z² − 5)^3/2 + C.

You can fix a missing constant factor this way, but not a missing variable. If the integrand were √(z² − 5) with no z outside, this u wouldn't work.

Example 3: ∫ x / √(x − 1) dx (solving for x)

Let u = x − 1, so du = dx. The leftover x in the numerator is written in terms of u: x = u + 1.
The integral becomes ∫ (u + 1) / u^1/2 du = ∫ (u^1/2 + u^−1/2) du.
Integrate: (2/3) u^3/2 + 2u^1/2 + C.
Factor and substitute back: (2/3) u^1/2(u + 3) = (2/3)√(x − 1)(x + 2) + C.

Definite integrals: change the limits

For a definite integral, convert the limits to u-values too, and you never have to substitute back:

∫_a^b f(g(x)) g′(x) dx = ∫_g(a)^g(b) f(u) du

Example 4: ∫₀¹ x²(1 + 2x³)⁵ dx

Let u = 1 + 2x³, so du = 6x² dx and x² dx = (1/6) du.
Change the limits: when x = 0, u = 1; when x = 1, u = 3.
The integral becomes (1/6) ∫₁³ u⁵ du = (1/6) · [u⁶/6] from 1 to 3 = (1/36)(729 − 1).
Answer: 728/36 = 182/9 ≈ 20.22.

One more: for ∫₀¹ x e^{4x² + 3} dx, let u = 4x² + 3 (limits 3 to 7) and x dx = (1/8) du. The result is (e⁷ − e³)/8 ≈ 134.568.

Common mistakes

Leaving some x in the integral after substituting.
Forgetting to convert dx to du, including the constant.
Changing the variable but keeping the old x-limits on a definite integral.
Dropping the + C on an indefinite integral.

Check answers with the integral calculator, approximate definite integrals with the Riemann sum calculator, or differentiate your answer with the derivative calculator to confirm you get the integrand back. For integrals with √(a² − x²) and similar, see trig substitution.

Source: OpenStax Calculus Volume 1, 5.5 Substitution.

Every worked example on this page was checked with Calcist's own engine and an independent reference.

Questions people ask

What is u-substitution?

A method for integrals that reverses the chain rule: set u equal to an inner function whose derivative also appears, rewrite the integral in u, and integrate.

How do I know what to choose for u?

Pick an inner expression, such as the inside of a power, root or exponent, whose derivative also appears in the integrand, up to a constant factor.

Do you change the limits in u-substitution?

Yes, for definite integrals: replace a and b with g(a) and g(b). Then you don't need to substitute back to x.

What if the derivative is off by a constant?

Multiply and divide by the constant. For example, if du = 2z dz but you have z dz, use z dz = (1/2) du.