6.8 u-substitution

Achievable AP Calculus AB

6. Integration

Our AP Calculus AB course is currently in development and is a work-in-progress.

u-substitution

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What you’ll learn:

How to use $u$ -substitution to simplify an integral
Using $u$ -substitution on indefinite and definite integrals

$u$ -substitution is essentially the reverse chain rule. You use it when one function is nested inside another, and the derivative of the inner function (or something close to it) also appears in the integrand. By temporarily switching to a new variable, you can simplify the integral and apply basic integration rules more directly.

Steps for $u$ -substitution

Choose $u = g (x)$ .

Pick the “inner” function. Look for something nested, or an expression whose derivative shows up elsewhere in the integrand. Avoid choosing $u = x$ , since that doesn’t simplify anything.

Differentiate $u$

Differentiate with respect to $x$ :

$\frac{d u}{d x} = g^{'} (x)$

Then rewrite it in differential form:

$d u = g^{'} (x) d x$

Rewrite the integral

Substitute $u$ and $d u$ into the integral. After this step, there should be no $x$ left in the integrand.

Integrate with respect to $u$ .

Apply the usual integration rules, treating $u$ as the variable.

Plug $g (x)$ back in.

For an indefinite integral, include the constant of integration $+ C$ .

Examples

$\int 2 x cos (x^{2}) d x$

Solution

The expression $x^{2}$ is inside the cosine, so choose

$u = x^{2}$

Differentiate:

$d u = 2 x d x$

That matches the $2 x d x$ in the integral, so rewrite in terms of $u$ :

$\int cos (u) d u$

$= sin (u)$

$= sin (x^{2}) + C$

AP tip:

Even if the derivative of your chosen $u$ isn’t a perfect match, you can often fix it by multiplying or dividing by a constant.

$\int x e^{x^{2}} d x$

Solution

(spoiler)

Again, $x^{2}$ is nested, so let

$u = x^{2}$

Then

$d u = 2 x d x$

But the integrand has only $x d x$ . Divide by $2$ :

$\frac{1}{2} d u = x d x$

Now rewrite the integral:

$\int e^{u} \cdot \frac{1}{2} d u$

$= \frac{1}{2} \int e^{u} d u$

$= \frac{1}{2} e^{u}$

$= \frac{1}{2} e^{x^{2}} + C$

AP tip:

For quotients, a common strategy is to choose $u$ as the denominator (or a nested part of it). That often produces a $g^{'} (x)$ factor that can cancel with what’s in the numerator.

$\int \frac{x - 1}{x ^{2} - 2 x + 3} d x$

Solution

(spoiler)

Let

$u = x^{2} - 2 x + 3$

Differentiate:

$d u = (2 x - 2) d x$

Factor out $2$ :

$d u = 2 (x - 1) d x$

Solve for $(x - 1) d x$ :

$\frac{1}{2} d u = (x - 1) d x$

Rewrite the integral:

$\int \frac{1}{u} \cdot \frac{1}{2} d u$

$= \frac{1}{2} \int \frac{1}{u} d u$

$= \frac{1}{2} ln (u)$

$= \frac{1}{2} ln (x^{2} - 2 x + 3) + C$

$\int \frac{2 - 3 + ln ( 4 x )}{x} d x$

Solution

(spoiler)

This is a quotient, but choosing $u = x$ won’t help. Instead, notice that $\frac{d}{d x} [ln (4 x)] = \frac{1}{x}$ , which appears in the integrand. Choose the nested expression:

$u = - 3 + ln (4 x)$

Then

$d u = \frac{1}{x} d x$

Rewrite the integral:

$2 \int u d u$

$= 2 \int u^{1/2} d u$

$= \frac{2 u ^{3/2}}{3/2}$

$= \frac{4}{3} u^{3/2}$

$= \frac{4}{3} (- 3 + ln (4 x))^{3/2} + C$

$\int tan (2 x) sec^{2} (2 x) d x$

Solution

(spoiler)

The integrand contains a function and its derivative pattern: the derivative of $tan (x)$ is $sec^{2} (x)$ . Because the angle is $2 x$ , choose

$u = tan (2 x)$

Differentiate:

$d u = 2 sec^{2} (2 x) d x$

Solve for $sec^{2} (2 x) d x$ :

$\frac{1}{2} d u = sec^{2} (2 x) d x$

Rewrite and integrate:

$\int u \cdot \frac{1}{2} d u$

$= \frac{1}{2} \cdot \frac{u ^{2}}{2}$

$= \frac{1}{4} tan^{2} (2 x) + C$

Alternatively, this substitution also works:

$u = sec (2 x)$

Then

$d u = 2 sec (2 x) tan (2 x) d x$

$\frac{1}{2} d u = sec (2 x) tan (2 x)$

And the integral rewritten is again

$\int u \cdot \frac{1}{2} d u$

But the final answer when replacing $u$ with $sec (2 x)$ is

$= \frac{1}{4} sec^{2} (2 x) + C$

Both answers are correct because of the Pythagorean identity $tan^{2} (x) + 1 = sec^{2} (x)$ . The two antiderivatives differ only by a constant.

As a bonus problem, turn the integrand into a function involving only $sin (2 x)$ and $cos (2 x)$ before making an appropriate $u$ -substitution.

Extra technique

Here’s a useful variation: sometimes you’ll do a small algebra step after choosing $u$ so you can rewrite everything in terms of $u$ .

$\int x x - 1 d x$

Solution

(spoiler)

The nested expression is $x - 1$ , so let

$u = x - 1$

Then

$d u = d x$

The remaining issue is the $x$ outside the square root. Rewrite $x$ in terms of $u$ by solving for $x$ :

$u + 1 = x$

Now rewrite the integral:

$\int ((u + 1) \cdot u) d u$

Distribute $u$ :

$\int (u u + u) d u$

$= \int (u^{3/2} + u^{1/2}) d u$

Integrate term-by-term:

$\frac{2}{5} u^{5/2} + \frac{2}{3} u^{3/2} + C$

$= \frac{2}{5} (x - 1)^{5/2} + \frac{2}{3} (x - 1)^{3/2} + C$

$\int \frac{x ^{3}}{x ^{2} + 1} d x$

Solution

(spoiler)

Let

$u = x^{2} + 1$

$d u = 2 x d x$

The numerator $x^{3}$ can be split as $x^{2} \cdot x$ . Rewrite each part in terms of $u$ .

From $u = x^{2} + 1$ :

$x^{2} = u - 1$

From $d u = 2 x d x$ :

$x d x = \frac{1}{2} d u$

$x^{3} d x = x^{2} \cdot x d x = (u - 1) \cdot \frac{1}{2} d u$

Now rewrite and simplify:

$\int \frac{x ^{3}}{x ^{2} + 1} d x$

$= \int \frac{1}{u} \cdot (u - 1) \frac{1}{2} \cdot d u$

$= \frac{1}{2} \int \frac{u - 1}{u} d u$

$= \frac{1}{2} (\int 1 d u - \int \frac{1}{u} d u)$

$= \frac{1}{2} (u - ln (u))$

$= \frac{1}{2} (x^{2} + 1 - ln (x^{2} + 1)) + C$

Definite integrals and changing bounds

When you use $u$ -substitution on a definite integral, you have two standard ways to finish the problem.

Option 1: Change the bounds

Convert the limits from $x$ -values to $u$ -values. Then you can evaluate the antiderivative in terms of $u$ without converting back to $x$ .

Option 2: Don’t change the bounds.

Convert back to $x$ after integrating, and then evaluate using the original $x$ -bounds.

Examples

$\int_{0}^{2} \frac{3 x}{( x ^{2} + 1 ) ^{2}} d x$

(spoiler)

Calculations are often simpler if you factor out constants first:

$3 \int_{0}^{2} \frac{x}{( x ^{2} + 1 ) ^{2}} d x$

The denominator is $(x^{2} + 1)^{2}$ . If you choose $u = (x^{2} + 1)^{2}$ , then $d u$ becomes more complicated than what appears in the numerator. Instead, choose the inner expression:

$u = x^{2} + 1$

Then

$d u = 2 x d x$

Solve for $x d x$ :

$\frac{1}{2} d u = x d x$

At this point, you can finish in either of two ways.

Option 1: Change the bounds

When $x = 0$ (the lower bound),

$u = 0^{2} + 1 = 1$

When $x = 2$ (the upper bound),

$u = 2^{2} + 1 = 5$

Rewrite the integral with $u$ -bounds:

$3 \int_{1}^{5} \frac{1}{u ^{2}} \cdot \frac{1}{2} d u$

$= \frac{3}{2} \int_{1}^{5} u^{- 2} d u$

$= \frac{3}{2} \cdot \frac{u ^{- 2 + 1}}{- 2 + 1}$

$= \frac{3}{2} (- \frac{1}{u})_{1}^{5}$

$= \frac{3}{2} [- \frac{1}{5} - (- \frac{1}{1})]$

$= \frac{6}{5}$

Option 2: Don’t change the bounds

Rewrite in terms of $u$ , but keep the original $x$ -bounds:

$3 \int_{0}^{2} \frac{1}{u ^{2}} \cdot \frac{1}{2} d u$

Integrate, then substitute $u = x^{2} + 1$ back in before evaluating from $0$ to $2$ :

$= \frac{3}{2} (- \frac{1}{u})$

$= \frac{3}{2} (- \frac{1}{x ^{2} + 1})_{0}^{2}$

$= \frac{3}{2} (- \frac{1}{2 ^{2} + 1} - (- \frac{1}{0 ^{2} + 1}))$

$= \frac{6}{5}$

$u$ -substitution overview

Reverse chain rule technique for integration
Used when integrand has a nested function and its derivative (or close)
Simplifies integrals by changing variables

Steps for $u$ -substitution

Choose $u = g (x)$ (inner function)
Differentiate: $d u = g^{'} (x) d x$
Rewrite integral in terms of $u$ and $d u$ (no $x$ left)
Integrate with respect to $u$
Substitute $g (x)$ back in (add $+ C$ for indefinite integrals)

Key substitution strategies

If $d u$ differs by a constant, adjust by multiplying/dividing
For quotients, often choose denominator or nested part as $u$
For products, look for a factor whose derivative appears elsewhere

Example highlights

$\int 2 x cos (x^{2}) d x$ : $u = x^{2}$ , $d u = 2 x d x$
$\int x e^{x^{2}} d x$ : $u = x^{2}$ , $d u = 2 x d x$ , adjust by $1/2$
$\int \frac{x - 1}{x ^{2} - 2 x + 3} d x$ : $u = x^{2} - 2 x + 3$ , $d u = 2 (x - 1) d x$
$\int \frac{2 - 3 + l n ( 4 x )}{x} d x$ : $u = - 3 + ln (4 x)$ , $d u = \frac{1}{x} d x$
$\int tan (2 x) sec^{2} (2 x) d x$ : $u = tan (2 x)$ or $u = sec (2 x)$ both work

Extra techniques

If $x$ remains after substitution, express $x$ in terms of $u$
- Example: $u = x - 1$ , so $x = u + 1$
For $\int \frac{x ^{3}}{x ^{2} + 1} d x$ , split $x^{3}$ as $x^{2} \cdot x$ and rewrite both in terms of $u$

Definite integrals with $u$ -substitution

Option 1: Change bounds to $u$ -values, evaluate antiderivative in $u$
Option 2: Keep $x$ -bounds, substitute back to $x$ before evaluating
Both methods yield the same final answer

General tips

Always check that all $x$ are replaced by $u$ before integrating
Adjust $d u$ as needed to match the integrand
For definite integrals, either change bounds or substitute back before evaluating

u-substitution

What you’ll learn:

How to use $u$ -substitution to simplify an integral
Using $u$ -substitution on indefinite and definite integrals

Steps for $u$ -substitution

Choose $u = g (x)$ .

Pick the “inner” function. Look for something nested, or an expression whose derivative shows up elsewhere in the integrand. Avoid choosing $u = x$ , since that doesn’t simplify anything.

Differentiate $u$

Differentiate with respect to $x$ :

$\frac{d u}{d x} = g^{'} (x)$

Then rewrite it in differential form:

$d u = g^{'} (x) d x$

Rewrite the integral

Substitute $u$ and $d u$ into the integral. After this step, there should be no $x$ left in the integrand.

Integrate with respect to $u$ .

Apply the usual integration rules, treating $u$ as the variable.

Plug $g (x)$ back in.

For an indefinite integral, include the constant of integration $+ C$ .

Examples

$\int 2 x cos (x^{2}) d x$

Solution

The expression $x^{2}$ is inside the cosine, so choose

$u = x^{2}$

Differentiate:

$d u = 2 x d x$

That matches the $2 x d x$ in the integral, so rewrite in terms of $u$ :

$\int cos (u) d u$

$= sin (u)$

$= sin (x^{2}) + C$

AP tip:

Even if the derivative of your chosen $u$ isn’t a perfect match, you can often fix it by multiplying or dividing by a constant.

$\int x e^{x^{2}} d x$

Solution

(spoiler)

Again, $x^{2}$ is nested, so let

$u = x^{2}$

Then

$d u = 2 x d x$

But the integrand has only $x d x$ . Divide by $2$ :

$\frac{1}{2} d u = x d x$

Now rewrite the integral:

$\int e^{u} \cdot \frac{1}{2} d u$

$= \frac{1}{2} \int e^{u} d u$

$= \frac{1}{2} e^{u}$

$= \frac{1}{2} e^{x^{2}} + C$

AP tip:

For quotients, a common strategy is to choose $u$ as the denominator (or a nested part of it). That often produces a $g^{'} (x)$ factor that can cancel with what’s in the numerator.

$\int \frac{x - 1}{x ^{2} - 2 x + 3} d x$

Solution

(spoiler)

Let

$u = x^{2} - 2 x + 3$

Differentiate:

$d u = (2 x - 2) d x$

Factor out $2$ :

$d u = 2 (x - 1) d x$

Solve for $(x - 1) d x$ :

$\frac{1}{2} d u = (x - 1) d x$

Rewrite the integral:

$\int \frac{1}{u} \cdot \frac{1}{2} d u$

$= \frac{1}{2} \int \frac{1}{u} d u$

$= \frac{1}{2} ln (u)$

$= \frac{1}{2} ln (x^{2} - 2 x + 3) + C$

$\int \frac{2 - 3 + ln ( 4 x )}{x} d x$

Solution

(spoiler)

This is a quotient, but choosing $u = x$ won’t help. Instead, notice that $\frac{d}{d x} [ln (4 x)] = \frac{1}{x}$ , which appears in the integrand. Choose the nested expression:

$u = - 3 + ln (4 x)$

Then

$d u = \frac{1}{x} d x$

Rewrite the integral:

$2 \int u d u$

$= 2 \int u^{1/2} d u$

$= \frac{2 u ^{3/2}}{3/2}$

$= \frac{4}{3} u^{3/2}$

$= \frac{4}{3} (- 3 + ln (4 x))^{3/2} + C$

$\int tan (2 x) sec^{2} (2 x) d x$

Solution

(spoiler)

The integrand contains a function and its derivative pattern: the derivative of $tan (x)$ is $sec^{2} (x)$ . Because the angle is $2 x$ , choose

$u = tan (2 x)$

Differentiate:

$d u = 2 sec^{2} (2 x) d x$

Solve for $sec^{2} (2 x) d x$ :

$\frac{1}{2} d u = sec^{2} (2 x) d x$

Rewrite and integrate:

$\int u \cdot \frac{1}{2} d u$

$= \frac{1}{2} \cdot \frac{u ^{2}}{2}$

$= \frac{1}{4} tan^{2} (2 x) + C$

Alternatively, this substitution also works:

$u = sec (2 x)$

Then

$d u = 2 sec (2 x) tan (2 x) d x$

$\frac{1}{2} d u = sec (2 x) tan (2 x)$

And the integral rewritten is again

$\int u \cdot \frac{1}{2} d u$

But the final answer when replacing $u$ with $sec (2 x)$ is

$= \frac{1}{4} sec^{2} (2 x) + C$

Both answers are correct because of the Pythagorean identity $tan^{2} (x) + 1 = sec^{2} (x)$ . The two antiderivatives differ only by a constant.

As a bonus problem, turn the integrand into a function involving only $sin (2 x)$ and $cos (2 x)$ before making an appropriate $u$ -substitution.

Extra technique

Here’s a useful variation: sometimes you’ll do a small algebra step after choosing $u$ so you can rewrite everything in terms of $u$ .

$\int x x - 1 d x$

Solution

(spoiler)

The nested expression is $x - 1$ , so let

$u = x - 1$

Then

$d u = d x$

The remaining issue is the $x$ outside the square root. Rewrite $x$ in terms of $u$ by solving for $x$ :

$u + 1 = x$

Now rewrite the integral:

$\int ((u + 1) \cdot u) d u$

Distribute $u$ :

$\int (u u + u) d u$

$= \int (u^{3/2} + u^{1/2}) d u$

Integrate term-by-term:

$\frac{2}{5} u^{5/2} + \frac{2}{3} u^{3/2} + C$

$= \frac{2}{5} (x - 1)^{5/2} + \frac{2}{3} (x - 1)^{3/2} + C$

$\int \frac{x ^{3}}{x ^{2} + 1} d x$

Solution

(spoiler)

Let

$u = x^{2} + 1$

$d u = 2 x d x$

The numerator $x^{3}$ can be split as $x^{2} \cdot x$ . Rewrite each part in terms of $u$ .

From $u = x^{2} + 1$ :

$x^{2} = u - 1$

From $d u = 2 x d x$ :

$x d x = \frac{1}{2} d u$

$x^{3} d x = x^{2} \cdot x d x = (u - 1) \cdot \frac{1}{2} d u$

Now rewrite and simplify:

$\int \frac{x ^{3}}{x ^{2} + 1} d x$

$= \int \frac{1}{u} \cdot (u - 1) \frac{1}{2} \cdot d u$

$= \frac{1}{2} \int \frac{u - 1}{u} d u$

$= \frac{1}{2} (\int 1 d u - \int \frac{1}{u} d u)$

$= \frac{1}{2} (u - ln (u))$

$= \frac{1}{2} (x^{2} + 1 - ln (x^{2} + 1)) + C$

Definite integrals and changing bounds

When you use $u$ -substitution on a definite integral, you have two standard ways to finish the problem.

Option 1: Change the bounds

Convert the limits from $x$ -values to $u$ -values. Then you can evaluate the antiderivative in terms of $u$ without converting back to $x$ .

Option 2: Don’t change the bounds.

Convert back to $x$ after integrating, and then evaluate using the original $x$ -bounds.

Examples

$\int_{0}^{2} \frac{3 x}{( x ^{2} + 1 ) ^{2}} d x$

(spoiler)

Calculations are often simpler if you factor out constants first:

$3 \int_{0}^{2} \frac{x}{( x ^{2} + 1 ) ^{2}} d x$

The denominator is $(x^{2} + 1)^{2}$ . If you choose $u = (x^{2} + 1)^{2}$ , then $d u$ becomes more complicated than what appears in the numerator. Instead, choose the inner expression:

$u = x^{2} + 1$

Then

$d u = 2 x d x$

Solve for $x d x$ :

$\frac{1}{2} d u = x d x$

At this point, you can finish in either of two ways.

Option 1: Change the bounds

When $x = 0$ (the lower bound),

$u = 0^{2} + 1 = 1$

When $x = 2$ (the upper bound),

$u = 2^{2} + 1 = 5$

Rewrite the integral with $u$ -bounds:

$3 \int_{1}^{5} \frac{1}{u ^{2}} \cdot \frac{1}{2} d u$

$= \frac{3}{2} \int_{1}^{5} u^{- 2} d u$

$= \frac{3}{2} \cdot \frac{u ^{- 2 + 1}}{- 2 + 1}$

$= \frac{3}{2} (- \frac{1}{u})_{1}^{5}$

$= \frac{3}{2} [- \frac{1}{5} - (- \frac{1}{1})]$

$= \frac{6}{5}$

Option 2: Don’t change the bounds

Rewrite in terms of $u$ , but keep the original $x$ -bounds:

$3 \int_{0}^{2} \frac{1}{u ^{2}} \cdot \frac{1}{2} d u$

Integrate, then substitute $u = x^{2} + 1$ back in before evaluating from $0$ to $2$ :

$= \frac{3}{2} (- \frac{1}{u})$

$= \frac{3}{2} (- \frac{1}{x ^{2} + 1})_{0}^{2}$

$= \frac{3}{2} (- \frac{1}{2 ^{2} + 1} - (- \frac{1}{0 ^{2} + 1}))$

$= \frac{6}{5}$

Achievable AP Calculus AB

6. Integration

Our AP Calculus AB course is currently in development and is a work-in-progress.

u-substitution

7 min read

Font

Discuss

Feedback

What you’ll learn:

How to use $u$ -substitution to simplify an integral
Using $u$ -substitution on indefinite and definite integrals

Steps for $u$ -substitution

Choose $u = g (x)$ .

Pick the “inner” function. Look for something nested, or an expression whose derivative shows up elsewhere in the integrand. Avoid choosing $u = x$ , since that doesn’t simplify anything.

Differentiate $u$

Differentiate with respect to $x$ :

$\frac{d u}{d x} = g^{'} (x)$

Then rewrite it in differential form:

$d u = g^{'} (x) d x$

Rewrite the integral

Substitute $u$ and $d u$ into the integral. After this step, there should be no $x$ left in the integrand.

Integrate with respect to $u$ .

Apply the usual integration rules, treating $u$ as the variable.

Plug $g (x)$ back in.

For an indefinite integral, include the constant of integration $+ C$ .

Examples

$\int 2 x cos (x^{2}) d x$

Solution

The expression $x^{2}$ is inside the cosine, so choose

$u = x^{2}$

Differentiate:

$d u = 2 x d x$

That matches the $2 x d x$ in the integral, so rewrite in terms of $u$ :

$\int cos (u) d u$

$= sin (u)$

$= sin (x^{2}) + C$

AP tip:

Even if the derivative of your chosen $u$ isn’t a perfect match, you can often fix it by multiplying or dividing by a constant.

$\int x e^{x^{2}} d x$

Solution

(spoiler)

Again, $x^{2}$ is nested, so let

$u = x^{2}$

Then

$d u = 2 x d x$

But the integrand has only $x d x$ . Divide by $2$ :

$\frac{1}{2} d u = x d x$

Now rewrite the integral:

$\int e^{u} \cdot \frac{1}{2} d u$

$= \frac{1}{2} \int e^{u} d u$

$= \frac{1}{2} e^{u}$

$= \frac{1}{2} e^{x^{2}} + C$

AP tip:

For quotients, a common strategy is to choose $u$ as the denominator (or a nested part of it). That often produces a $g^{'} (x)$ factor that can cancel with what’s in the numerator.

$\int \frac{x - 1}{x ^{2} - 2 x + 3} d x$

Solution

(spoiler)

Let

$u = x^{2} - 2 x + 3$

Differentiate:

$d u = (2 x - 2) d x$

Factor out $2$ :

$d u = 2 (x - 1) d x$

Solve for $(x - 1) d x$ :

$\frac{1}{2} d u = (x - 1) d x$

Rewrite the integral:

$\int \frac{1}{u} \cdot \frac{1}{2} d u$

$= \frac{1}{2} \int \frac{1}{u} d u$

$= \frac{1}{2} ln (u)$

$= \frac{1}{2} ln (x^{2} - 2 x + 3) + C$

$\int \frac{2 - 3 + ln ( 4 x )}{x} d x$

Solution

(spoiler)

This is a quotient, but choosing $u = x$ won’t help. Instead, notice that $\frac{d}{d x} [ln (4 x)] = \frac{1}{x}$ , which appears in the integrand. Choose the nested expression:

$u = - 3 + ln (4 x)$

Then

$d u = \frac{1}{x} d x$

Rewrite the integral:

$2 \int u d u$

$= 2 \int u^{1/2} d u$

$= \frac{2 u ^{3/2}}{3/2}$

$= \frac{4}{3} u^{3/2}$

$= \frac{4}{3} (- 3 + ln (4 x))^{3/2} + C$

$\int tan (2 x) sec^{2} (2 x) d x$

Solution

(spoiler)

The integrand contains a function and its derivative pattern: the derivative of $tan (x)$ is $sec^{2} (x)$ . Because the angle is $2 x$ , choose

$u = tan (2 x)$

Differentiate:

$d u = 2 sec^{2} (2 x) d x$

Solve for $sec^{2} (2 x) d x$ :

$\frac{1}{2} d u = sec^{2} (2 x) d x$

Rewrite and integrate:

$\int u \cdot \frac{1}{2} d u$

$= \frac{1}{2} \cdot \frac{u ^{2}}{2}$

$= \frac{1}{4} tan^{2} (2 x) + C$

Alternatively, this substitution also works:

$u = sec (2 x)$

Then

$d u = 2 sec (2 x) tan (2 x) d x$

$\frac{1}{2} d u = sec (2 x) tan (2 x)$

And the integral rewritten is again

$\int u \cdot \frac{1}{2} d u$

But the final answer when replacing $u$ with $sec (2 x)$ is

$= \frac{1}{4} sec^{2} (2 x) + C$

Both answers are correct because of the Pythagorean identity $tan^{2} (x) + 1 = sec^{2} (x)$ . The two antiderivatives differ only by a constant.

As a bonus problem, turn the integrand into a function involving only $sin (2 x)$ and $cos (2 x)$ before making an appropriate $u$ -substitution.

Extra technique

Here’s a useful variation: sometimes you’ll do a small algebra step after choosing $u$ so you can rewrite everything in terms of $u$ .

$\int x x - 1 d x$

Solution

(spoiler)

The nested expression is $x - 1$ , so let

$u = x - 1$

Then

$d u = d x$

The remaining issue is the $x$ outside the square root. Rewrite $x$ in terms of $u$ by solving for $x$ :

$u + 1 = x$

Now rewrite the integral:

$\int ((u + 1) \cdot u) d u$

Distribute $u$ :

$\int (u u + u) d u$

$= \int (u^{3/2} + u^{1/2}) d u$

Integrate term-by-term:

$\frac{2}{5} u^{5/2} + \frac{2}{3} u^{3/2} + C$

$= \frac{2}{5} (x - 1)^{5/2} + \frac{2}{3} (x - 1)^{3/2} + C$

$\int \frac{x ^{3}}{x ^{2} + 1} d x$

Solution

(spoiler)

Let

$u = x^{2} + 1$

$d u = 2 x d x$

The numerator $x^{3}$ can be split as $x^{2} \cdot x$ . Rewrite each part in terms of $u$ .

From $u = x^{2} + 1$ :

$x^{2} = u - 1$

From $d u = 2 x d x$ :

$x d x = \frac{1}{2} d u$

$x^{3} d x = x^{2} \cdot x d x = (u - 1) \cdot \frac{1}{2} d u$

Now rewrite and simplify:

$\int \frac{x ^{3}}{x ^{2} + 1} d x$

$= \int \frac{1}{u} \cdot (u - 1) \frac{1}{2} \cdot d u$

$= \frac{1}{2} \int \frac{u - 1}{u} d u$

$= \frac{1}{2} (\int 1 d u - \int \frac{1}{u} d u)$

$= \frac{1}{2} (u - ln (u))$

$= \frac{1}{2} (x^{2} + 1 - ln (x^{2} + 1)) + C$

Definite integrals and changing bounds

When you use $u$ -substitution on a definite integral, you have two standard ways to finish the problem.

Option 1: Change the bounds

Convert the limits from $x$ -values to $u$ -values. Then you can evaluate the antiderivative in terms of $u$ without converting back to $x$ .

Option 2: Don’t change the bounds.

Convert back to $x$ after integrating, and then evaluate using the original $x$ -bounds.

Examples

$\int_{0}^{2} \frac{3 x}{( x ^{2} + 1 ) ^{2}} d x$

(spoiler)

Calculations are often simpler if you factor out constants first:

$3 \int_{0}^{2} \frac{x}{( x ^{2} + 1 ) ^{2}} d x$

The denominator is $(x^{2} + 1)^{2}$ . If you choose $u = (x^{2} + 1)^{2}$ , then $d u$ becomes more complicated than what appears in the numerator. Instead, choose the inner expression:

$u = x^{2} + 1$

Then

$d u = 2 x d x$

Solve for $x d x$ :

$\frac{1}{2} d u = x d x$

At this point, you can finish in either of two ways.

Option 1: Change the bounds

When $x = 0$ (the lower bound),

$u = 0^{2} + 1 = 1$

When $x = 2$ (the upper bound),

$u = 2^{2} + 1 = 5$

Rewrite the integral with $u$ -bounds:

$3 \int_{1}^{5} \frac{1}{u ^{2}} \cdot \frac{1}{2} d u$

$= \frac{3}{2} \int_{1}^{5} u^{- 2} d u$

$= \frac{3}{2} \cdot \frac{u ^{- 2 + 1}}{- 2 + 1}$

$= \frac{3}{2} (- \frac{1}{u})_{1}^{5}$

$= \frac{3}{2} [- \frac{1}{5} - (- \frac{1}{1})]$

$= \frac{6}{5}$

Option 2: Don’t change the bounds

Rewrite in terms of $u$ , but keep the original $x$ -bounds:

$3 \int_{0}^{2} \frac{1}{u ^{2}} \cdot \frac{1}{2} d u$

Integrate, then substitute $u = x^{2} + 1$ back in before evaluating from $0$ to $2$ :

$= \frac{3}{2} (- \frac{1}{u})$

$= \frac{3}{2} (- \frac{1}{x ^{2} + 1})_{0}^{2}$

$= \frac{3}{2} (- \frac{1}{2 ^{2} + 1} - (- \frac{1}{0 ^{2} + 1}))$

$= \frac{6}{5}$

$u$ -substitution overview

Reverse chain rule technique for integration
Used when integrand has a nested function and its derivative (or close)
Simplifies integrals by changing variables

Steps for $u$ -substitution

Choose $u = g (x)$ (inner function)
Differentiate: $d u = g^{'} (x) d x$
Rewrite integral in terms of $u$ and $d u$ (no $x$ left)
Integrate with respect to $u$
Substitute $g (x)$ back in (add $+ C$ for indefinite integrals)

Key substitution strategies

If $d u$ differs by a constant, adjust by multiplying/dividing
For quotients, often choose denominator or nested part as $u$
For products, look for a factor whose derivative appears elsewhere

Example highlights

$\int 2 x cos (x^{2}) d x$ : $u = x^{2}$ , $d u = 2 x d x$
$\int x e^{x^{2}} d x$ : $u = x^{2}$ , $d u = 2 x d x$ , adjust by $1/2$
$\int \frac{x - 1}{x ^{2} - 2 x + 3} d x$ : $u = x^{2} - 2 x + 3$ , $d u = 2 (x - 1) d x$
$\int \frac{2 - 3 + l n ( 4 x )}{x} d x$ : $u = - 3 + ln (4 x)$ , $d u = \frac{1}{x} d x$
$\int tan (2 x) sec^{2} (2 x) d x$ : $u = tan (2 x)$ or $u = sec (2 x)$ both work

Extra techniques

If $x$ remains after substitution, express $x$ in terms of $u$
- Example: $u = x - 1$ , so $x = u + 1$
For $\int \frac{x ^{3}}{x ^{2} + 1} d x$ , split $x^{3}$ as $x^{2} \cdot x$ and rewrite both in terms of $u$

Definite integrals with $u$ -substitution

Option 1: Change bounds to $u$ -values, evaluate antiderivative in $u$
Option 2: Keep $x$ -bounds, substitute back to $x$ before evaluating
Both methods yield the same final answer

General tips

Always check that all $x$ are replaced by $u$ before integrating
Adjust $d u$ as needed to match the integrand
For definite integrals, either change bounds or substitute back before evaluating

u-substitution

What you’ll learn:

How to use $u$ -substitution to simplify an integral
Using $u$ -substitution on indefinite and definite integrals

Steps for $u$ -substitution

Choose $u = g (x)$ .

Pick the “inner” function. Look for something nested, or an expression whose derivative shows up elsewhere in the integrand. Avoid choosing $u = x$ , since that doesn’t simplify anything.

Differentiate $u$

Differentiate with respect to $x$ :

$\frac{d u}{d x} = g^{'} (x)$

Then rewrite it in differential form:

$d u = g^{'} (x) d x$

Rewrite the integral

Substitute $u$ and $d u$ into the integral. After this step, there should be no $x$ left in the integrand.

Integrate with respect to $u$ .

Apply the usual integration rules, treating $u$ as the variable.

Plug $g (x)$ back in.

For an indefinite integral, include the constant of integration $+ C$ .

Examples

$\int 2 x cos (x^{2}) d x$

Solution

The expression $x^{2}$ is inside the cosine, so choose

$u = x^{2}$

Differentiate:

$d u = 2 x d x$

That matches the $2 x d x$ in the integral, so rewrite in terms of $u$ :

$\int cos (u) d u$

$= sin (u)$

$= sin (x^{2}) + C$

AP tip:

Even if the derivative of your chosen $u$ isn’t a perfect match, you can often fix it by multiplying or dividing by a constant.

$\int x e^{x^{2}} d x$

Solution

(spoiler)

Again, $x^{2}$ is nested, so let

$u = x^{2}$

Then

$d u = 2 x d x$

But the integrand has only $x d x$ . Divide by $2$ :

$\frac{1}{2} d u = x d x$

Now rewrite the integral:

$\int e^{u} \cdot \frac{1}{2} d u$

$= \frac{1}{2} \int e^{u} d u$

$= \frac{1}{2} e^{u}$

$= \frac{1}{2} e^{x^{2}} + C$

AP tip:

For quotients, a common strategy is to choose $u$ as the denominator (or a nested part of it). That often produces a $g^{'} (x)$ factor that can cancel with what’s in the numerator.

$\int \frac{x - 1}{x ^{2} - 2 x + 3} d x$

Solution

(spoiler)

Let

$u = x^{2} - 2 x + 3$

Differentiate:

$d u = (2 x - 2) d x$

Factor out $2$ :

$d u = 2 (x - 1) d x$

Solve for $(x - 1) d x$ :

$\frac{1}{2} d u = (x - 1) d x$

Rewrite the integral:

$\int \frac{1}{u} \cdot \frac{1}{2} d u$

$= \frac{1}{2} \int \frac{1}{u} d u$

$= \frac{1}{2} ln (u)$

$= \frac{1}{2} ln (x^{2} - 2 x + 3) + C$

$\int \frac{2 - 3 + ln ( 4 x )}{x} d x$

Solution

(spoiler)

This is a quotient, but choosing $u = x$ won’t help. Instead, notice that $\frac{d}{d x} [ln (4 x)] = \frac{1}{x}$ , which appears in the integrand. Choose the nested expression:

$u = - 3 + ln (4 x)$

Then

$d u = \frac{1}{x} d x$

Rewrite the integral:

$2 \int u d u$

$= 2 \int u^{1/2} d u$

$= \frac{2 u ^{3/2}}{3/2}$

$= \frac{4}{3} u^{3/2}$

$= \frac{4}{3} (- 3 + ln (4 x))^{3/2} + C$

$\int tan (2 x) sec^{2} (2 x) d x$

Solution

(spoiler)

The integrand contains a function and its derivative pattern: the derivative of $tan (x)$ is $sec^{2} (x)$ . Because the angle is $2 x$ , choose

$u = tan (2 x)$

Differentiate:

$d u = 2 sec^{2} (2 x) d x$

Solve for $sec^{2} (2 x) d x$ :

$\frac{1}{2} d u = sec^{2} (2 x) d x$

Rewrite and integrate:

$\int u \cdot \frac{1}{2} d u$

$= \frac{1}{2} \cdot \frac{u ^{2}}{2}$

$= \frac{1}{4} tan^{2} (2 x) + C$

Alternatively, this substitution also works:

$u = sec (2 x)$

Then

$d u = 2 sec (2 x) tan (2 x) d x$

$\frac{1}{2} d u = sec (2 x) tan (2 x)$

And the integral rewritten is again

$\int u \cdot \frac{1}{2} d u$

But the final answer when replacing $u$ with $sec (2 x)$ is

$= \frac{1}{4} sec^{2} (2 x) + C$

Both answers are correct because of the Pythagorean identity $tan^{2} (x) + 1 = sec^{2} (x)$ . The two antiderivatives differ only by a constant.

As a bonus problem, turn the integrand into a function involving only $sin (2 x)$ and $cos (2 x)$ before making an appropriate $u$ -substitution.

Extra technique

Here’s a useful variation: sometimes you’ll do a small algebra step after choosing $u$ so you can rewrite everything in terms of $u$ .

$\int x x - 1 d x$

Solution

(spoiler)

The nested expression is $x - 1$ , so let

$u = x - 1$

Then

$d u = d x$

The remaining issue is the $x$ outside the square root. Rewrite $x$ in terms of $u$ by solving for $x$ :

$u + 1 = x$

Now rewrite the integral:

$\int ((u + 1) \cdot u) d u$

Distribute $u$ :

$\int (u u + u) d u$

$= \int (u^{3/2} + u^{1/2}) d u$

Integrate term-by-term:

$\frac{2}{5} u^{5/2} + \frac{2}{3} u^{3/2} + C$

$= \frac{2}{5} (x - 1)^{5/2} + \frac{2}{3} (x - 1)^{3/2} + C$

$\int \frac{x ^{3}}{x ^{2} + 1} d x$

Solution

(spoiler)

Let

$u = x^{2} + 1$

$d u = 2 x d x$

The numerator $x^{3}$ can be split as $x^{2} \cdot x$ . Rewrite each part in terms of $u$ .

From $u = x^{2} + 1$ :

$x^{2} = u - 1$

From $d u = 2 x d x$ :

$x d x = \frac{1}{2} d u$

$x^{3} d x = x^{2} \cdot x d x = (u - 1) \cdot \frac{1}{2} d u$

Now rewrite and simplify:

$\int \frac{x ^{3}}{x ^{2} + 1} d x$

$= \int \frac{1}{u} \cdot (u - 1) \frac{1}{2} \cdot d u$

$= \frac{1}{2} \int \frac{u - 1}{u} d u$

$= \frac{1}{2} (\int 1 d u - \int \frac{1}{u} d u)$

$= \frac{1}{2} (u - ln (u))$

$= \frac{1}{2} (x^{2} + 1 - ln (x^{2} + 1)) + C$

Definite integrals and changing bounds

When you use $u$ -substitution on a definite integral, you have two standard ways to finish the problem.

Option 1: Change the bounds

Convert the limits from $x$ -values to $u$ -values. Then you can evaluate the antiderivative in terms of $u$ without converting back to $x$ .

Option 2: Don’t change the bounds.

Convert back to $x$ after integrating, and then evaluate using the original $x$ -bounds.

Examples

$\int_{0}^{2} \frac{3 x}{( x ^{2} + 1 ) ^{2}} d x$

(spoiler)

Calculations are often simpler if you factor out constants first:

$3 \int_{0}^{2} \frac{x}{( x ^{2} + 1 ) ^{2}} d x$

The denominator is $(x^{2} + 1)^{2}$ . If you choose $u = (x^{2} + 1)^{2}$ , then $d u$ becomes more complicated than what appears in the numerator. Instead, choose the inner expression:

$u = x^{2} + 1$

Then

$d u = 2 x d x$

Solve for $x d x$ :

$\frac{1}{2} d u = x d x$

At this point, you can finish in either of two ways.

Option 1: Change the bounds

When $x = 0$ (the lower bound),

$u = 0^{2} + 1 = 1$

When $x = 2$ (the upper bound),

$u = 2^{2} + 1 = 5$

Rewrite the integral with $u$ -bounds:

$3 \int_{1}^{5} \frac{1}{u ^{2}} \cdot \frac{1}{2} d u$

$= \frac{3}{2} \int_{1}^{5} u^{- 2} d u$

$= \frac{3}{2} \cdot \frac{u ^{- 2 + 1}}{- 2 + 1}$

$= \frac{3}{2} (- \frac{1}{u})_{1}^{5}$

$= \frac{3}{2} [- \frac{1}{5} - (- \frac{1}{1})]$

$= \frac{6}{5}$

Option 2: Don’t change the bounds

Rewrite in terms of $u$ , but keep the original $x$ -bounds:

$3 \int_{0}^{2} \frac{1}{u ^{2}} \cdot \frac{1}{2} d u$

Integrate, then substitute $u = x^{2} + 1$ back in before evaluating from $0$ to $2$ :

$= \frac{3}{2} (- \frac{1}{u})$

$= \frac{3}{2} (- \frac{1}{x ^{2} + 1})_{0}^{2}$

$= \frac{3}{2} (- \frac{1}{2 ^{2} + 1} - (- \frac{1}{0 ^{2} + 1}))$

$= \frac{6}{5}$

u-substitution

What you’ll learn:

Steps for u-substitution

Examples

Solution

Solution

Solution

Solution

Solution

Extra technique

Solution

Solution

Definite integrals and changing bounds

Examples

u-substitution

What you’ll learn:

Steps for u-substitution

Examples

Solution

Solution

Solution

Solution

Solution

Extra technique

Solution

Solution

Definite integrals and changing bounds

Examples

u-substitution

What you’ll learn:

Steps for u-substitution

Examples

Solution

Solution

Solution

Solution

Solution

Extra technique

Solution

Solution

Definite integrals and changing bounds

Examples

u-substitution

What you’ll learn:

Steps for u-substitution

Examples

Solution

Solution

Solution

Solution

Solution

Extra technique

Solution

Solution

Definite integrals and changing bounds

Examples

Steps for $u$ -substitution

Steps for $u$ -substitution

Steps for $u$ -substitution

Steps for $u$ -substitution