Unit 6: Integration & Accumulation of Change

U-Substitution

Finding antiderivatives using u-substitution.

Advanced25 min lesson3 min readUpdated August 12, 2026Author not yet attributed

Prerequisites

Applying the Fundamental Theorem

Undoing the Chain Rule

∫ 2x(x² + 1)⁴ dx doesn't match any basic antiderivative pattern directly — but it looks suspiciously like the result of a chain rule differentiation, run backwards. Before reading on: if (x² + 1)⁵ were differentiated, what factor would the chain rule contribute, and does it look familiar here?

Definition — U-Substitution

U-substitution reverses the chain rule for integration: choosing u equal to an inner function, computing du, and rewriting the entire integral in terms of u alone (leaving no x behind) before integrating. Once the antiderivative is found in terms of u, substitute the original expression for u back in.

\int f(g(x))\,g'(x)\,dx = \int f(u)\,du, \quad u = g(x)

Worked Example — An Indefinite Integral by Substitution

Evaluate ∫ 2x(x² + 1)⁴ dx. Let u = x² + 1, so du = 2x dx — which matches the 2x dx factor in the integral exactly. Rewrite entirely in u: ∫ u⁴ du = u⁵/5 + C. Substitute back: (x² + 1)⁵/5 + C. Check by differentiating: d/dx[(x²+1)⁵/5] = (1/5)(5)(x²+1)⁴(2x) = 2x(x²+1)⁴ — matching the original integrand exactly.

Worked Example — A Definite Integral, Converting the Bounds

Evaluate ∫₀² x√(x² + 4) dx. Let u = x² + 4, so du = 2x dx, meaning x dx = du/2. Convert the bounds to u as well, rather than converting back to x at the end: when x = 0, u = 4; when x = 2, u = 8. The integral becomes ∫₄⁸ √u · (du/2) = (1/2)∫₄⁸ u^(1/2) du = (1/2) · [(2/3)u^(3/2)]₄⁸ = (1/3)[u^(3/2)]₄⁸ = (1/3)(8^(3/2) − 4^(3/2)). Since 8^(3/2) = √(8³) = √512 = 16√2 and 4^(3/2) = √(4³) = √64 = 8: (1/3)(16√2 − 8) ≈ (1/3)(22.627 − 8) ≈ 4.876.

Tip

For a definite integral, converting the bounds to u-values (as shown above) avoids ever having to substitute back to x — evaluate the entire antiderivative in u, at the u-bounds, and stop there.

Common Mistakes

Leaving a leftover x in the integral after substituting for u, and integrating with respect to u anyway.
A correct substitution eliminates every x from the integral — if an x remains after substituting, either du doesn't match what's available in the integral, or a different u needs to be chosen.
Using the original x-bounds after converting a definite integral to u, instead of converting the bounds themselves.
Once the integral is rewritten entirely in u, its bounds must be rewritten in u too — reusing the original x-bounds evaluates the u-antiderivative at the wrong points entirely.

Key Takeaways

U-substitution reverses the chain rule: choosing u as an inner function whose derivative (du) is available elsewhere in the integral, then integrating entirely in terms of u.
A correct substitution leaves no x behind — every remaining piece of the integral must be expressible in u.
For a definite integral, converting the bounds of integration to u-values avoids ever substituting back to x.

Summary

This closes Unit 6: Riemann sums, the definite integral, the Fundamental Theorem, and u-substitution together let accumulated change be found exactly. Unit 7 turns integration toward a new kind of equation — one that specifies a rate of change and asks for the original function.

Track your progress

← Applying the Fundamental Theorem Back to Unit 6: Integration & Accumulation of Change