How do I know what to pick for u?

Choose the inner function whose derivative (up to a constant multiple) also appears in the integrand. Common picks are the expression inside a power or root, the exponent of e, the argument of a trig function, or a denominator. If your choice makes every remaining x disappear after substitution, you picked well.

Do I have to change the limits for definite integrals?

No, but you must be consistent. If you change the limits to u-values, evaluate the antiderivative in u and never convert back. If you keep the original x-limits, you must first convert the antiderivative back to x before plugging in. Changing the limits is usually faster and less error-prone.

What if the derivative of my inner function isn't in the integrand?

First check whether it is off only by a constant—constants can be adjusted freely. If a variable factor is left over, try solving your substitution for x and rewriting that factor in terms of u, as in xsqrt(x+1)\,dx. If nothing works, the integral likely needs a different technique such as integration by parts.

Why did I lose points even though my answer looked right?

The most common causes are forgetting the differential factor (e.g., missing a 13 from du = 3\,dx), omitting +C on an indefinite integral, or leaving x-limits on an antiderivative written in u. Always double-check that every x and dx was replaced and that your bounds match your variable.

AP-CALCBC-6.9

U6.9 Integration by Substitution

Master u-substitution for AP Calculus BC: identify the inner function, substitute, integrate, convert back, and change limits for definite integrals.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on U6.9 Integration by Substitution, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start U6.9 Integration by Substitution free

What this lesson covers

Integration by substitution reverses the chain rule. Whenever you see a composite function multiplied by (a constant times) the derivative of its inner function, u-substitution unlocks the integral. It is the single most-used integration technique on the AP exam, and mastering it now sets you up for integration by parts and partial fractions later.

In this lesson you will learn how to spot a good substitution, carry it out cleanly, and handle the special step definite integrals require: changing the limits of integration. We will also cover the most common mistakes—forgetting

du

, mismatched constants, and reverting incorrectly—so you can avoid them under time pressure.

Why Substitution Works

The chain rule says

\frac{d}{dx}\big[F(g(x))\big] = F'(g(x))\cdot g'(x)

. Integration by substitution simply runs this backward. If an integrand looks like

f(g(x))\cdot g'(x)

, then letting

u = g(x)

turns the whole expression into

f(u)\,du

, which is usually a basic antiderivative.

The key move is the differential. If

u = g(x)

, then

\frac{du}{dx} = g'(x)

, so

du = g'(x)\,dx

. You are literally trading the

x

-world for the

u

-world. Every

x

(including the

dx

) must disappear and be replaced by

u

and

du

before you integrate.

A common misconception is thinking you need the derivative of the inner function to appear exactly. In fact you only need it up to a constant multiple, because constants can be factored in and out of integrals. For example, if

du = 2x\,dx

but the integrand only has

x\,dx

, you write

x\,dx = \tfrac{1}{2}\,du

and continue. You cannot, however, fix a mismatch involving a variable factor—if the leftover cannot be written entirely in terms of

u

, substitution alone will not finish the job.

The Step-by-Step Procedure

Follow the same reliable routine every time.

Step	Action
1	Choose $u$ = the inner function (often inside a power, root, exponent, or denominator)
2	Compute $du = g'(x)\,dx$
3	Solve for $dx$ or for the grouped factor, and substitute everything
4	Integrate in terms of $u$
5	Replace $u$ with $g(x)$ to return to $x$ (indefinite integrals)
6	Add $+C$

How do you pick

u

? Look for a function whose derivative also appears (up to a constant) in the integrand. Good candidates: the expression under a radical, the exponent of

e

, the argument of a trig function, or a denominator.

Consider

\int 2x\cos(x^2)\,dx

. Let

u = x^2

, so

du = 2x\,dx

. The integral becomes

\int \cos(u)\,du = \sin(u) + C = \sin(x^2) + C

. Notice how the

2x\,dx

was perfectly consumed by

du

. The AP exam loves integrands engineered exactly this way, so training your eye to match derivatives is the whole skill.

Definite Integrals: Changing the Limits

For a definite integral you have two valid options, and choosing well saves time.

Option A—change the limits. When you substitute

u = g(x)

, convert the

x

-limits into

u

-limits using

u = g(x)

. Then you never convert back; you evaluate entirely in

u

. This is the AP-preferred, error-resistant method.

Option B—keep the original limits, but then you MUST convert

u

back to

x

before plugging in the

x

-bounds. A fatal error is leaving

x

-limits on a

u

-antiderivative.

Example with changed limits:

\int_{0}^{2} x\,e^{x^2}\,dx

. Let

u = x^2

du = 2x\,dx

, so

x\,dx = \tfrac{1}{2}\,du

. When

x=0

u=0

; when

x=2

u=4

. The integral becomes

\tfrac{1}{2}\int_{0}^{4} e^{u}\,du = \tfrac{1}{2}\big[e^{u}\big]_0^4 = \tfrac{1}{2}(e^4 - 1)

	Change limits	Keep limits
Convert back to $x$ ?	No	Yes
Bounds used	$u$ -values	original $x$ -values
Risk	Forgetting to convert bounds	Forgetting to convert antiderivative

On free-response questions, always label your new limits clearly. Graders want to see that the bounds correspond to whichever variable your antiderivative is written in.

Common Traps and Exam Tips

The most frequent error is dropping the differential. Writing

u = 3x+1

then integrating

\int (3x+1)^5\,dx

as if it were

\int u^5

ignores that

du = 3\,dx

, so

dx = \tfrac{1}{3}\,du

. The correct answer carries a factor of

\tfrac{1}{3}

.

A second trap is a leftover variable. In

\int x\sqrt{x+1}\,dx

, letting

u = x+1

gives

du = dx

, but the stray

x

must be rewritten as

x = u - 1

. Then

\int (u-1)\sqrt{u}\,du

expands into integrable powers. Do not abandon substitution just because the derivative is not sitting there—sometimes solving for

x

rescues the problem.

Third, watch constant handling. You can pull constants out (

\int 5f\,du = 5\int f\,du

) but never pull out variables.

For multiple choice, working backward by differentiating each answer choice with the chain rule is a fast check. For calculator-active questions, you can substitute and then numerically integrate. On no-calculator sections, expect substitutions that produce clean logs, exponentials, or standard trig antiderivatives. Recognizing

\int \frac{g'(x)}{g(x)}\,dx = \ln|g(x)| + C

as a substitution pattern is especially high-yield.

Recognizing Substitution Patterns Quickly

Speed on the AP exam comes from pattern recognition. Train yourself to associate integrand shapes with the substitution that clears them.

Integrand shape	Choose $u$	Result type
$[g(x)]^n\,g'(x)$	$u = g(x)$	power rule $\frac{u^{n+1}}{n+1}$
$e^{g(x)}\,g'(x)$	$u = g(x)$	$e^{u}$
$\frac{g'(x)}{g(x)}$	$u = g(x)$	$\ln\|u\|$
$\cos(g(x))\,g'(x)$	$u = g(x)$	$\sin(u)$
$\frac{g'(x)}{1+[g(x)]^2}$	$u = g(x)$	$\arctan(u)$

The unifying idea: find a chunk whose derivative is also present. Radicals suggest letting

u

equal what is inside. Fractions often reward setting

u

equal to the denominator and checking whether the numerator is (a multiple of) its derivative.

If a first choice of

u

does not clear all the

x

's, do not panic—either solve for

x

in terms of

u

to eliminate a leftover, or reconsider your choice. When substitution genuinely fails (no derivative match and no algebraic fix), that signals a different technique from later lessons, such as integration by parts or partial fractions. Building this decision instinct is exactly what U6.14 will formalize.

Key terms

u-substitution.: An integration technique that reverses the chain rule by replacing an inner function $g(x)$ with a new variable $u$ , turning a composite integrand into a simpler one.
Inner function.: The function $g(x)$ nested inside another function; it is the natural choice for $u$ because its derivative typically appears in the integrand.
Differential.: The expression $du = g'(x)\,dx$ , which converts the $dx$ portion of the integral into $du$ during substitution.
Composite function.: A function of the form $f(g(x))$ , where one function is applied to the output of another; substitution targets these.
Changing the limits.: For definite integrals, converting the original $x$ -bounds into $u$ -bounds via $u = g(x)$ so the integral can be evaluated entirely in $u$ .
Integration constant.: The $+C$ added to every indefinite integral to account for the family of antiderivatives; omitted only after applying definite bounds.

Worked example

Evaluate

\int_{1}^{e} \frac{\ln x}{x}\,dx

Identify the inner function. The integrand contains

\ln x

and its derivative

\frac{1}{x}

, which is a classic substitution signal. Let

u = \ln x

.

Compute the differential:

du = \frac{1}{x}\,dx

. Notice the

\frac{1}{x}\,dx

already sitting in the integral becomes exactly

du

, and

\ln x

becomes

u

.

Change the limits (preferred method). When

x = 1

u = \ln 1 = 0

. When

x = e

u = \ln e = 1

. So the new bounds are from

u=0

u=1

.

Rewrite the integral entirely in

u

\int_{0}^{1} u\,du

.

Integrate:

\int u\,du = \frac{u^2}{2}

. Because we changed the limits, we evaluate in

u

directly and do not convert back.

Apply the bounds:

\left[\frac{u^2}{2}\right]_0^1 = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2}

.

The value of the definite integral is

\frac{1}{2}

. As a check, note the antiderivative in

x

\frac{(\ln x)^2}{2}

, and evaluating from 1 to

e

gives

\frac{1}{2} - 0 = \frac{1}{2}

—consistent.

Practice questions

Evaluate

\int 6x^2\sqrt{x^3+4}\,dx

$\frac{4}{3}(x^3+4)^{3/2} + C$
$\frac{2}{3}(x^3+4)^{3/2} + C$
$(x^3+4)^{3/2} + C$
$\frac{9}{2}(x^3+4)^{3/2} + C$

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

Let

u = x^3 + 4

, so

du = 3x^2\,dx

, giving

6x^2\,dx = 2\,du

. The integral becomes

\int 2\sqrt{u}\,du = 2\cdot \frac{2}{3}u^{3/2} = \frac{4}{3}u^{3/2}

. Substituting back yields

\frac{4}{3}(x^3+4)^{3/2} + C

. The trap answers come from mishandling the constant factor from

du

Evaluate

\int \frac{x}{\sqrt{x-3}}\,dx

using substitution, showing how to handle the leftover variable.

Answer: $\frac{2}{3}(x-3)^{3/2} + 6(x-3)^{1/2} + C$

Let

u = x - 3

, so

du = dx

and

x = u + 3

. The integral becomes

\int \frac{u+3}{\sqrt{u}}\,du = \int \left(u^{1/2} + 3u^{-1/2}\right)du = \frac{2}{3}u^{3/2} + 6u^{1/2} + C

. Converting back gives

\frac{2}{3}(x-3)^{3/2} + 6(x-3)^{1/2} + C

. This problem tests whether you can rescue a substitution by solving for

x

when a stray variable remains.

Evaluate

\int_{0}^{\pi/2} \cos x\, e^{\sin x}\,dx

Answer: $e - 1$

Let

u = \sin x

, so

du = \cos x\,dx

. Change limits: when

x=0

u=0

; when

x=\pi/2

u=1

. The integral becomes

\int_0^1 e^u\,du = [e^u]_0^1 = e^1 - e^0 = e - 1

. Changing the limits avoids converting the antiderivative back to

x

FAQ

How do I know what to pick for u?: Choose the inner function whose derivative (up to a constant multiple) also appears in the integrand. Common picks are the expression inside a power or root, the exponent of $e$ , the argument of a trig function, or a denominator. If your choice makes every remaining $x$ disappear after substitution, you picked well.
Do I have to change the limits for definite integrals?: No, but you must be consistent. If you change the limits to $u$ -values, evaluate the antiderivative in $u$ and never convert back. If you keep the original $x$ -limits, you must first convert the antiderivative back to $x$ before plugging in. Changing the limits is usually faster and less error-prone.
What if the derivative of my inner function isn't in the integrand?: First check whether it is off only by a constant—constants can be adjusted freely. If a variable factor is left over, try solving your substitution for $x$ and rewriting that factor in terms of $u$ , as in $\int x\sqrt{x+1}\,dx$ . If nothing works, the integral likely needs a different technique such as integration by parts.
Why did I lose points even though my answer looked right?: The most common causes are forgetting the differential factor (e.g., missing a $\tfrac{1}{3}$ from $du = 3\,dx$ ), omitting $+C$ on an indefinite integral, or leaving $x$ -limits on an antiderivative written in $u$ . Always double-check that every $x$ and $dx$ was replaced and that your bounds match your variable.

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