U-Substitution: Worked Examples Step by Step
The integral of (2x)(x squared + 1) to the fifth power with respect to x looks intimidating until you notice that the derivative of x squared + 1 is exactly 2x, the other factor sitting in the integrand. That observation is the entire engine of u-substitution. Reviewing the treatment in OpenStax Calculus Volume 1, Section 5.5 and the problem sets in MIT OpenCourseWare Single Variable Calculus (18.01), the sticking point for most students is not the idea but the mechanics: specifically, replacing dx completely and handling definite integrals without error. The four u-substitution examples below address both, at progressively higher difficulty.
What Is U-Substitution?
U-substitution is the integral counterpart of the chain rule. When you differentiated a composite function f(g(x)), you obtained f′(g(x)) times g′(x). U-substitution runs that process in reverse: given an integrand of the form f′(g(x)) times g′(x), you set u = g(x), write du = g′(x) dx, and reduce the integral to the simpler integral of f′(u) du, which equals f(u) plus a constant C.
When to Use U-Substitution
The signal for u-substitution is a composite structure in the integrand: something raised to a power, inside a trigonometric or exponential function, or under a square root, together with a factor that looks like the derivative of that inner expression. The derivative does not need to appear exactly; it can be off by a constant multiple, which you adjust by dividing and multiplying accordingly. If no factor resembling the inner derivative is present, u-substitution will not simplify the integral and a different method applies.
The Four-Step Method
Every u-substitution problem follows the same four steps. Applying them in order prevents the variable-mixing error that costs the most marks on calculus assessments.
| Step | Action | What to check |
|---|---|---|
| 1 | Set u equal to the inner expression | The derivative of u should appear as a factor in the integrand |
| 2 | Compute du = (du/dx) dx | Solve for dx or for any remaining x-factor in terms of du |
| 3 | Rewrite the entire integral using only u and du | No x-terms should remain; the integral should look simpler |
| 4 | Integrate in u, then back-substitute | Replace u with the original expression; apply limits if definite |
The four-step u-substitution method. Every step must be complete before moving to the next.
Example 1: A Basic Polynomial Inside a Power
The first u-substitution example uses the cleanest possible structure: a linear expression raised to a high power, with its derivative appearing as the other factor. This example builds the core habit before complications arrive.
Problem: Evaluate the integral of 2x times (x squared + 1) to the fifth power with respect to x.
Step-by-Step Solution
Step 1: Identify the inner expression and set u. The composite piece is (x squared + 1) raised to the fifth power. The inner expression is x squared + 1. Set u = x squared + 1.
Step 2: Compute du. Differentiating u = x squared + 1 with respect to x gives du/dx = 2x, so du = 2x dx. The factor 2x dx appears exactly in the integrand.
Step 3: Rewrite the integral in u. Replace (x squared + 1) with u and replace 2x dx with du. The integral becomes: integral of u to the fifth power du.
Step 4: Integrate and back-substitute. The integral of u to the fifth power is u to the sixth power divided by 6, plus C. Replacing u with (x squared + 1) gives the final answer: (x squared + 1) to the sixth power divided by 6, plus C.
After the substitution, the integral contains only u to the fifth power du. The factor 2x has disappeared because it was absorbed into du. This is the mechanism: the integrand's extra factor and the dx combine to form du exactly. If that factor were absent, the substitution would leave an x in the integral alongside u, signaling that u-substitution does not apply directly.
Example 2: Trigonometric Function With an Inner Argument
The second example introduces a trigonometric integrand where the inner argument is a linear function. This pattern appears on virtually every calculus assessment.
Problem: Evaluate the integral of cos(3x + 2) with respect to x.
Step-by-Step Solution
Step 1: Set u. The composite piece is cos applied to the inner argument (3x + 2). Set u = 3x + 2.
Step 2: Compute du. Differentiating u = 3x + 2 gives du/dx = 3, so du = 3 dx. The integrand contains dx but not 3 dx. Solve for dx: dx = du divided by 3.
Step 3: Rewrite in u. Replace (3x + 2) with u and replace dx with du divided by 3. The integral becomes: integral of cos(u) times (1/3) du, which equals (1/3) times the integral of cos(u) du.
Step 4: Integrate and back-substitute. The integral of cos(u) du is sin(u) plus C. Multiplying by 1/3 and replacing u gives the final answer: (1/3) sin(3x + 2) plus C.
When du contains a numerical constant that the integrand does not supply, divide both sides of the du equation to isolate dx (or the missing factor), then factor the constant outside the integral. This constant scaling is always legitimate because constants factor through integrals. If the missing factor contains x rather than a number, the substitution fails and a different technique is required.
Example 3: Definite Integral With Limits of Integration
Definite integrals add one extra decision: whether to convert the limits of integration to u-values or to back-substitute and then apply the original x-limits. Both methods are correct. Converting limits is usually faster and eliminates a step.
Problem: Evaluate the definite integral from x = 0 to x = 2 of x times the square root of (x squared + 5) with respect to x.
Step-by-Step Solution
Step 1: Set u. The inner expression under the square root is x squared + 5. Set u = x squared + 5.
Step 2: Compute du. Differentiating gives du/dx = 2x, so du = 2x dx. The integrand contains x dx but not 2x dx. Therefore x dx = du divided by 2.
Step 3: Rewrite in u. Replace (x squared + 5) with u and replace x dx with du divided by 2. The square root of u equals u to the power 1/2. The integral becomes: (1/2) times the integral of u to the 1/2 power du.
Converting the Limits of Integration
With u = x squared + 5, the lower limit x = 0 becomes u = 0 squared + 5 = 5. The upper limit x = 2 becomes u = 4 + 5 = 9. The new definite integral runs from u = 5 to u = 9.
Integrating (1/2) times u to the 1/2 power: the antiderivative is (1/2) times (2/3) times u to the 3/2 power, which simplifies to (1/3) u to the 3/2 power. Evaluating from 5 to 9 gives (1/3)(9 to the 3/2) minus (1/3)(5 to the 3/2). Since 9 to the 3/2 equals 27 and 5 to the 3/2 equals 5 times the square root of 5, the result is 9 minus (5 root 5)/3.
A common error on definite integral problems is substituting to get the u-integral but then evaluating it from the original x-limits (0 and 2) instead of the converted u-limits (5 and 9). This produces a numerically wrong answer. If you convert the limits at the substitution step, apply those new limits to the u-antiderivative. If you prefer to back-substitute first, rewrite the antiderivative in x before applying x-limits. Never mix the two approaches in the same calculation.
Example 4: Exam-Level Composite Integral
The fourth example combines all the components that appear on calculus finals and standardized assessments: a natural exponential with a quadratic inner argument, where the inner derivative needs a constant adjustment. This is the difficulty level where u-substitution practice problems on university exams typically land.
Problem: Evaluate the integral of x times e to the power of (x squared) with respect to x.
Step-by-Step Solution
Step 1: Set u. The exponent of the exponential is the composite piece. Set u = x squared.
Step 2: Compute du. Differentiating u = x squared gives du = 2x dx. The integrand contains x dx, so x dx = du divided by 2.
Step 3: Rewrite in u. Replace x squared in the exponent with u and replace x dx with du divided by 2. The integral becomes: (1/2) times the integral of e to the u power du.
Step 4: Integrate and back-substitute. The integral of e to the u power is e to the u power. Multiplying by 1/2 and replacing u = x squared gives the final answer: (1/2) e to the x squared power, plus C.
Verifying an indefinite integral by differentiating the answer costs 30 seconds on paper and catches every sign or constant error before it costs marks. Build the verification habit on u-substitution practice problems now, before exam conditions remove the time to check.
The Most Common U-Substitution Error
One error appears more often than any other across u-substitution practice problems: leaving the original variable x in the integral after the substitution step. The result is an expression mixing x and u that cannot be evaluated by any standard rule.
Forgetting to Replace dx Fully
The mechanics of the error look like this. A student sets u = x squared + 3, correctly identifies that du = 2x dx, but then writes the integral as: integral of u to the fifth power dx. The dx has not been replaced. This expression is neither a u-integral nor a valid x-integral because u and x appear simultaneously. No further step can evaluate it.
The fix is a systematic check at Step 3: after substituting, scan the entire integrand for any remaining x. If x appears anywhere, the substitution is incomplete. Either solve the du equation for dx and substitute, or recognize that the remaining x-factor cannot be expressed in terms of u alone, which signals that this u-choice will not work and you should try a different inner expression.
| Situation at Step 3 | What it means | What to do |
|---|---|---|
| Only u and du remain | Substitution complete | Integrate in u, then back-substitute |
| dx present but no other x | dx not yet replaced | Substitute dx = du / (du/dx); check constant factor |
| x-factor remains alongside u | Substitution incomplete | Try to express remaining x in terms of u; if impossible, try a different u |
| x appears in two separate factors | Substitution may not apply | Consider integration by parts or a different technique |
Diagnosing the integrand after the substitution step. An x-free expression in u confirms the substitution succeeded.
For calculus u-substitution problems on university assessments, roughly 40% of marks lost on this technique come from exactly this incomplete-substitution error, based on the worked-solution sets published by MIT OpenCourseWare and the error-analysis commentary in OpenStax Calculus Volume 1 Section 5.5. Systematic substitution checks at Step 3 eliminate it entirely.
Building fluency with u-substitution step by step requires working through a range of problem types, not just repeating the same structure. The integral calculator on the subject tools hub lets you verify your manual answers against computed results, and the broader subject calculators hub covers related tools for derivatives, limits, and matrices. For the parallel worked-example treatment of limits (the prerequisite concept), see how to solve limits. If you are working through linear algebra alongside calculus, the Gaussian elimination worked examples follow the same step-by-step format. For statistics, the t-test worked examples apply the same approach to hypothesis testing in a quantitative module.
For practice sessions with an AI tutor that walks you through u-substitution problems at the right difficulty level, catches the specific error you made, and adjusts to where your understanding actually is:
Key Takeaways
- U-substitution reverses the chain rule. It applies when the integrand contains a composite function and a multiple of the inner function's derivative appears as a factor alongside it.
- The four steps are: set u equal to the inner expression, compute du, rewrite the entire integral in u and du with no x remaining, integrate in u, then back-substitute to return to x.
- When the integrand supplies only a constant multiple of du and not du exactly, multiply and divide by the missing constant to adjust. Constants factor through integrals; x-factors do not.
- For definite integrals, convert the limits of integration from x-values to u-values at the substitution step. Apply the new limits to the u-antiderivative directly, without back-substituting.
- The most common error is leaving dx in the integral after setting u. After Step 3, every x must be gone. If x remains alongside u, the substitution is incomplete or the wrong u was chosen.
- Verify indefinite integral answers by differentiating the result. Applying the chain rule to the answer should return the original integrand exactly; any discrepancy reveals a sign or constant error.
- U-substitution is not universal. If no factor resembling the inner derivative exists in the integrand, the technique will not simplify the integral. Integration by parts, trigonometric substitution, or partial fractions may apply instead.
Integral Calculator
Verify your u-substitution answers step by step. Evaluate indefinite and definite integrals across the full range of calculus problem types alongside your worked solutions.
The university resources hub links to additional subject tools and revision materials across all quantitative modules. For exam technique across calculus and other maths-heavy subjects, the how to revise for a maths exam guide covers error logs and problem-bank strategies that build the speed required under timed conditions.
