U-substitution is the integration equivalent of the chain rule — and it's the most common method you'll use beyond the basic power rule. Here's how to master it.
The Core Idea
If the integrand contains a composite function AND pieces of its derivative, let u = the inner function and rewrite everything in terms of u.
∫f(g(x)) · g'(x) dx → ∫f(u) du
The 4-Step Process
- Choose u — pick the "inner" function
- Find du — differentiate u to get du/dx, then solve for dx
- Substitute — replace all x-expressions with u-expressions
- Integrate and back-substitute — solve in u, then convert back to x
Worked Examples
Example 1: ∫2x(x²+3)⁴ dx
- u = x² + 3
- du = 2x dx → the 2x dx is already in the integrand!
- ∫u⁴ du = u⁵/5 + C
- = (x²+3)⁵/5 + C
Example 2: ∫cos(3x) · 3 dx
- u = 3x
- du = 3 dx → matches the 3 in the integrand
- ∫cos(u) du = sin(u) + C
- = sin(3x) + C
Example 3: ∫xe^(x²) dx
- u = x²
- du = 2x dx → we have x dx, not 2x dx
- Adjust: (1/2)du = x dx
- (1/2)∫eᵘ du = (1/2)eᵘ + C
- = (1/2)e^(x²) + C
Example 4: ∫x²/√(x³+1) dx
- u = x³ + 1
- du = 3x² dx → (1/3)du = x² dx
- (1/3)∫u^(−1/2) du = (1/3) · 2u^(1/2) + C
- = (2/3)√u + C
- = (2/3)√(x³+1) + C
How to Choose u
| Good choice for u | Why | |-------------------|-----| | Inside of a composite function | Chain rule in reverse | | Under a radical | Simplifies when squared | | In an exponent | Turns exponential into algebraic | | In the denominator | Often simplifies the fraction |
Bad choice: The outer function. If you see sin(3x), u = 3x (inner), NOT u = sin(3x).
The "du Check" — Does Substitution Work?
After choosing u, check if du (or a constant multiple of du) appears in the integrand.
| Integrand | u | du | du present? | Works? | |-----------|---|----|-------------|--------| | 2x(x²+1)³ dx | x²+1 | 2x dx | ✅ Yes | ✅ | | x·cos(x²) dx | x² | 2x dx | ⚠️ x dx = (1/2)du | ✅ (adjust) | | sin(x²) dx | x² | 2x dx | ❌ No x outside | ❌ | | e^(3x) dx | 3x | 3 dx | ⚠️ dx = (1/3)du | ✅ (adjust) |
Common U-Substitution Mistakes
| Mistake | Wrong | Right | |---------|-------|-------| | Forgetting to substitute dx | ∫u⁴ dx | ∫u⁴ du | | Not adjusting for constant | du = 2x dx, only have x dx | (1/2)∫u⁴ du | | Forgetting to back-substitute | u⁵/5 + C | (x²+3)⁵/5 + C | | Choosing wrong u | u = x² in ∫x²·eˣ dx | u = x² doesn't work — use by parts |
The Trench Truth: If u-substitution doesn't work after 30 seconds, it's probably the wrong method. Switch to integration by parts or partial fractions. Don't burn 10 minutes forcing a substitution that was never going to fit.
Try our integral calculator to verify your u-substitution work with full step-by-step solutions.
Related: Derivative Calculator · Quadratic Formula Calculator · Square Root Calculator
Discussion
Loading comments...