Integration by Substitution

What You’ll Learn

In this lesson you’ll learn u-substitution, the most common integration technique. It’s the chain rule in reverse.

The Concept

When you learned the chain rule for derivatives, you saw that the derivative of a composite function like (x² + 5)⁴ involves differentiating the outside and multiplying by the derivative of the inside:

$$\frac{d}{dx}(x^2 + 5)^4 = 4(x^2 + 5)^3 \cdot 2x$$

U-substitution is this process in reverse. If you see an integral that looks like it came from a chain rule derivative, you can undo it.

The idea: temporarily replace the complicated inside part with a single variable u. This turns a hard integral into an easy one.

Why it works

Consider the integral:

$$\int 2x(x^2 + 5)^3 \, dx$$

This looks messy. But notice: x² + 5 is inside the cube, and 2x (the derivative of x² + 5) is sitting right there as a factor. If you let u = x² + 5, then du = 2x dx. The entire integral collapses to:

$$\int u^3 \, du$$

That’s just the power rule. You integrate, get u⁴/4 + C, then swap u back for x² + 5.

The procedure

1. Choose u = (the inside function, the part that’s nested inside something else)
2. Compute du = (derivative of u) dx
3. Check: does du (or a constant multiple of it) appear in the integrand?
4. Rewrite the entire integral in terms of u and du. No x’s should remain.
5. Integrate with respect to u using the basic rules
6. Substitute back: replace u with the original expression in x

What if du doesn’t match exactly?

Sometimes the derivative of u is off by a constant factor. For example, if u = 3x, then du = 3 dx, but you only have dx in the integral. That’s fine: just solve for dx = du/3 and adjust with the constant. You can always pull constant factors in or out. What you can’t do is pull x’s in or out.

The general formula:

$$\int f(g(x)) \cdot g'(x) \, dx = \int f(u) \, du \quad \text{where } u = g(x)$$

How to spot a good substitution

Ask yourself: “Is there a function inside another function? Is the derivative of the inside function (or something close to it) also in the integrand?” If yes, let u be the inside function.

Worked Example

Example 1: Power of a function

Evaluate:

$$\int 2x(x^2 + 5)^3 \, dx$$

Let u = x² + 5. Then du = 2x dx. The 2x dx in the integrand is exactly du.

$$\int u^3 \, du = \frac{u^4}{4} + C = \frac{(x^2 + 5)^4}{4} + C$$

Example 2: Trig with chain rule

Evaluate:

$$\int \sin(3x) \, dx$$

Let u = 3x. Then du = 3 dx, so dx = du/3.

$$\int \sin(u) \cdot \frac{du}{3} = -\frac{1}{3}\cos(u) + C = -\frac{1}{3}\cos(3x) + C$$

The 1/3 factor comes from adjusting for the chain rule. When the inside function has a constant multiplier, you divide by that constant.

Example 3: Square root with substitution

Evaluate:

$$\int x\sqrt{x^2 + 1} \, dx$$

Let u = x² + 1. Then du = 2x dx, so x dx = du/2.

$$\frac{1}{2}\int \sqrt{u} \, du = \frac{1}{2} \cdot \frac{2}{3}u^{3/2} + C = \frac{1}{3}(x^2 + 1)^{3/2} + C$$

Real-World Application

U-substitution shows up whenever you integrate a composite function:

• Physics: integrating force along a curved path where the path variable needs substitution
• Economics: computing total revenue when the rate function involves a composition
• Biology: integrating growth models where the rate depends on a transformed variable
• Engineering: computing work done by a variable force

You’ve Got This

U-substitution is pattern recognition. Look for something inside a function whose derivative is floating around in the integrand. If you see x² + 1 inside a power or root, and 2x (or a multiple of it) is also there, that’s your cue. Choose u = the inside, compute du, rewrite, integrate, substitute back. Same five steps every time.

Quiz

U-substitution is the reverse of:

• A.The power rule
• B.The product rule
• C.The chain rule
• D.The quotient rule

For the integral of 2x(x² + 5)³ dx, the best choice for u is:

• A.2x
• B.x² + 5
• C.(x² + 5)³
• D.x

The integral of sin(3x) dx equals:

• A.cos(3x) + C
• B.3cos(3x) + C
• C.-(1/3)cos(3x) + C
• D.-3cos(3x) + C

After integrating with respect to u, you must:

• A.Take the derivative
• B.Add limits of integration
• C.Substitute back to the original variable
• D.Multiply by 2

A good u is usually:

• A.The entire integrand
• B.A constant
• C.The outside function
• D.The inside of a composite function