How do I know which part of the integral to use as u?

Look for a composite function (an inner function inside another function) and check if its derivative appears somewhere in the integrand. The inner function is usually your best choice for u. For example, in \int 2x(x^2+1)^5 dx, the inner function is x^2+1 and its derivative 2x is visible in front, so you choose u = x^2+1.

Why do the limits of integration change when I use U-substitution on a definite integral?

When you replace x with u = g(x), the variable of integration changes, so the limits must change too. If you integrate from x=a to x=b, you now integrate from u=g(a) to u=g(b). This ensures the integral represents the same area or quantity before and after substitution.

What should I do if du does not match exactly what appears in my integral?

Pull out constants and scale appropriately. If du = 4x \, dx but the integrand shows $x \, dx$, rewrite x \, dx = \frac{1}{4} du. If the integrand shows $3x \, dx$, then 3x \, dx = \frac{3}{4} du. Factor out the coefficient in front of the integral sign and continue with the simpler form.

Can I always use U-substitution to solve any integral?

No. U-substitution requires that the derivative of your chosen u (or a simple multiple of it) appears in the integrand. If this condition is not met, the substitution will not simplify the integral. In those cases, try Integration by Parts, Partial Fractions, trigonometric substitution, or look up a table of antiderivatives.

U-Substitution

U-substitution simplifies integrals by transforming them into easier forms using a change of variables, ideal when you spot a function and its derivative.

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

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What each symbol means

What U-Substitution takes

Symbol	Meaning
$f$	The outer function in a composite $f(g(x))$ that you are integrating; if you choose $u$ correctly, the integral becomes $\int f(u) \, du$ , which is simpler to solve.
$g$	The inner function in a composite expression; its derivative $g'(x)$ must appear in the original integrand for U-substitution to work cleanly and avoid messy algebra.
$u$	The new variable that replaces $g(x)$ during substitution; the integral is rewritten entirely in terms of $u$ , often becoming a standard form like a power or exponential.
$x$	The original variable of integration; U-substitution converts the integral from $\int f(g(x))g'(x) \, dx$ to $\int f(u) \, du$ , and you substitute $x$ back at the end.

When to use it

Use U-substitution when an integral contains a composite function and its derivative appears in the integrand.

Level

Usually taught in: Calculus I · Appears on: AP Calculus

Worked examples

1. Integrate a polynomial composite function

Problem

Find

\int (x^2 + 1)^5 \cdot 2x \, dx

$u = x^2 + 1, \quad du = 2x \, dx$
Identify the inside function $u = x^2 + 1$ and differentiate it to get $du = 2x \, dx$ , which matches exactly what appears in the integrand.
$\int (x^2 + 1)^5 \cdot 2x \, dx = \int u^5 \, du$
Substitute the new variable $u$ and $du$ to replace the composite expression and rewrite the integral in a simpler form.
$\int u^5 \, du = \frac{u^6}{6} + C$
Apply the power rule for integration: $\int u^n \, du = \frac{u^{n+1}}{n+1} + C$ with $n = 5$ .
$\frac{(x^2+1)^6}{6} + C$
Substitute $u = x^2 + 1$ back into the antiderivative to express the final answer in terms of the original variable $x$ .

Answer: $\frac{(x^2+1)^6}{6} + C$

This is a textbook case for U-substitution: the integrand contains a composite function raised to a power, and its derivative is multiplied directly in front. Recognizing that $u = x^2 + 1$ and $du = 2x \, dx$ lets us turn the integral into a simple power rule, which would be tedious to expand by hand.

2. Integrate a rational function with a negative exponent

Problem

Find

\int \frac{-6x}{(2x^2-5)^3} \, dx

$u = 2x^2 - 5, \quad du = 4x \, dx$
Let $u$ be the denominator base, and differentiate: $du = 4x \, dx$ .
$-6x \, dx = -\frac{3}{2} \, du$
From $du = 4x \, dx$ , we have $x \, dx = \frac{1}{4} du$ , so $-6x \, dx = -6 \cdot \frac{1}{4} du = -\frac{3}{2} du$ . This step is crucial: the coefficient must match.
$\int \frac{-6x}{(2x^2-5)^3} \, dx = \int \frac{-3/2}{u^3} \, du = -\frac{3}{2} \int u^{-3} \, du$
Substitute to get a simple power integral with a negative exponent.
$-\frac{3}{2} \cdot \frac{u^{-2}}{-2} + C = \frac{3}{4} u^{-2} + C$
Apply the power rule: $\int u^{-3} \, du = \frac{u^{-2}}{-2}$ , then simplify by multiplying $-\frac{3}{2}$ by $\frac{1}{-2}$ .
$\frac{3}{4(2x^2-5)^2} + C$
Substitute $u = 2x^2 - 5$ back and write the answer with positive exponents.

Answer: $\frac{3}{4(2x^2-5)^2} + C$

This example requires careful bookkeeping: because $du = 4x \, dx$ but the integral shows $-6x \, dx$ , we must account for the coefficient ratio. The negative exponent and fraction also make this harder to spot at first, but U-substitution still cuts through the algebra cleanly.

3. Model fuel consumption on a road trip

Problem

A car on a road trip consumes fuel at a rate of

r(t) = 120e^{-0.5t}

gallons per hour, where

t

is time in hours. Due to the driver settling into a steady pace, consumption decreases exponentially. How much fuel was consumed during the first

2

hours?

$\int_0^2 120e^{-0.5t} \, dt$
Set up the integral of the consumption rate to find total fuel used.
$u = -0.5t, \quad du = -0.5 \, dt, \quad dt = -2 \, du$
Let $u$ be the exponent, then rearrange to express $dt$ in terms of $du$ .
$\text{When } t = 0: u = 0; \quad \text{When } t = 2: u = -1$
Convert the limits of integration to the new variable: substituting $t=0$ gives $u=0$ , and $t=2$ gives $u=-1$ .
$\int_0^2 120e^{-0.5t} \, dt = \int_0^{-1} 120e^u \cdot (-2) \, du = -240 \int_0^{-1} e^u \, du$
Substitute $u$ , $du$ , and the new limits into the integral.
$-240 [e^u]_0^{-1} = -240(e^{-1} - e^0) = -240(e^{-1} - 1)$
Integrate $e^u$ and apply the Fundamental Theorem of Calculus.
$-240e^{-1} + 240 = 240 - 240e^{-1}$
Distribute the $-240$ and rearrange to get the final form: $240(1 - e^{-1})$ gallons.

Answer: $240 - 240e^{-1} \text{ gallons} \approx 151.7 \text{ gallons}$

This problem shows U-substitution in a real-world setting: modeling changing rates. The exponential decay in fuel consumption is realistic, and the integral $\int e^{ax} \, dx$ always requires U-substitution (or knowledge of a formula). Changing the limits when substituting is essential for definite integrals, and this example shows why: the limits for $t$ are different from those for $u$ .

Common mistakes

Where U-Substitution usually goes wrong

The mistake	Why it is wrong	The fix
Writing $\int u^5 \, du = \frac{u^6}{6} + C$ and stopping, leaving the answer in terms of $u$ instead of $x$ .	The problem asks for an antiderivative in terms of the original variable $x$ ; an answer in terms of $u$ is incomplete and incorrect.	Always substitute your original variable back in at the end: $\frac{u^6}{6} + C = \frac{(x^2+1)^6}{6} + C$ .
If $u = x^2 + 1$ , incorrectly writing $du = 2x^2 \, dx$ instead of $du = 2x \, dx$ .	The derivative of $x^2$ is $2x$ , not $2x^2$ . Using the wrong derivative will lead to an incorrect constant factor in the answer.	Carefully differentiate: $\frac{d}{dx}(x^2 + 1) = 2x$ , so $du = 2x \, dx$ . Double-check this step before substituting.
In $\int \frac{-6x}{(2x^2-5)^3} dx$ with $u = 2x^2 - 5$ and $du = 4x \, dx$ , writing $\int \frac{-6}{u^3} du$ instead of $\int \frac{-3/2}{u^3} du$ .	When you substitute, all parts of $dx$ must be accounted for. Since $du = 4x \, dx$ , then $-6x \, dx = -\frac{3}{2} du$ , not $-6 du$ . Forgetting to adjust the coefficient causes the answer to be off by a factor.	After setting $u$ and $du$ , solve for $x \, dx$ (or whatever expression appears in the integral): $x \, dx = \frac{1}{4} du$ , then $-6x \, dx = -\frac{3}{2} du$ .

Tips and when to use something else

U-substitution works when the derivative of an inner function appears in the integrand; if it does not, try Integration by Parts or another method instead.
Always write down $du$ immediately after choosing $u$ , and check that $du$ (or a simple multiple of it) appears in the original integral—if not, your choice of $u$ may be wrong.
For definite integrals, change the limits of integration to match the new variable; this avoids having to convert back to the original variable and then re-evaluate.
If $du = 4x \, dx$ but your integral shows only $x \, dx$, divide both sides: $x \, dx = \frac{1}{4} du$ . Pull out fractional coefficients as constants in front of the integral sign.

Related formulas

Indefinite IntegralThe indefinite integral finds all functions whose derivative equals a given function, expressed as an antiderivative plus an arbitrary constant.Integration by PartsIntegration by Parts is a calculus technique for evaluating integrals of products by transforming them into simpler forms when substitution fails.Power Rule for IntegrationPower Rule for Integration instantly finds antiderivatives of polynomial terms; use it to integrate any power of x except x to the negative one.Definite Integral PropertiesTwo key properties of definite integrals: reversing the bounds negates the integral, and integrating over a single point gives zero.Fundamental Theorem of CalculusCompute the exact value of a definite integral using the antiderivative—connect area under a curve to the accumulation of change.

Frequently asked questions

How do I know which part of the integral to use as $u$ ?: Look for a composite function (an inner function inside another function) and check if its derivative appears somewhere in the integrand. The inner function is usually your best choice for $u$ . For example, in $\int 2x(x^2+1)^5 dx$ , the inner function is $x^2+1$ and its derivative $2x$ is visible in front, so you choose $u = x^2+1$ .
Why do the limits of integration change when I use U-substitution on a definite integral?: When you replace $x$ with $u = g(x)$ , the variable of integration changes, so the limits must change too. If you integrate from $x=a$ to $x=b$ , you now integrate from $u=g(a)$ to $u=g(b)$ . This ensures the integral represents the same area or quantity before and after substitution.
What should I do if $du$ does not match exactly what appears in my integral?: Pull out constants and scale appropriately. If $du = 4x \, dx$ but the integrand shows $x \, dx$, rewrite $x \, dx = \frac{1}{4} du$ . If the integrand shows $3x \, dx$, then $3x \, dx = \frac{3}{4} du$ . Factor out the coefficient in front of the integral sign and continue with the simpler form.
Can I always use U-substitution to solve any integral?: No. U-substitution requires that the derivative of your chosen $u$ (or a simple multiple of it) appears in the integrand. If this condition is not met, the substitution will not simplify the integral. In those cases, try Integration by Parts, Partial Fractions, trigonometric substitution, or look up a table of antiderivatives.

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Reviewed 2026-09-18