Blog · Calculus guides · 21 October 2026 · 4 min read

U-Substitution: Steps, Examples & Practice Problems

Q: Is u-substitution the reverse of the chain rule?

Yes. Differentiate ((x^2+1)^5)/5 with the chain rule and you get back 2x(x^2+1)^4. That is also the quickest way to check any substitution answer.

Q: Can you use u-substitution on definite integrals?

Yes. Change the limits as well: if u = x^2+1 and x runs from 0 to 1, then u runs from 1 to 2. Then you never have to substitute back.

U-substitution integrates a function whose inside and the inside's derivative both appear in the integrand. Set $u$ equal to the inside, replace $d x$ using $d u = u^{'} (x) d x$ , integrate the simpler expression in $u$ , then put $x$ back. For example, with $u = x^{2} + 1$ , $\int 2 x (x^{2} + 1)^{4} d x = \frac{( x ^{2} + 1 ) ^{5}}{5} + C$ .

The formula

\int f (g (x)) g^{'} (x) d x = \int f (u) d u, u = g (x), d u = g^{'} (x) d x

The chain rule, reversed. The chain rule multiplies by the inside derivative; substitution looks for that factor already sitting there and quietly removes it.

When to use it

Use substitution when you can see a function inside another function and the derivative of that inside is also in the integrand, up to a constant: $x cos (x^{2})$ has $x^{2}$ inside and $x$ (half its derivative) outside; $\frac{l n x}{x}$ has $ln x$ and its derivative $\frac{1}{x}$ . It is always worth trying before integration by parts, because when it works it is much shorter. If the inside's derivative is nowhere to be found, substitution usually turns the problem into a mess, and that is your sign to try something else.

How to do it, step by step

Choose $u$ . Pick the inside of the most awkward-looking piece: the bracket under a power, the argument of $sin$ or $e^{□}$ , the thing under a root, the denominator.
Find $d u$ . Differentiate: $d u = u^{'} (x) d x$ . Rearrange so you can swap out whatever is in the integrand, for example $x d x = \frac{1}{2} d u$ .
Rewrite the whole integral in $u$ . No $x$ may survive, and no $d x$ . If an $x$ is left over, solve $u = g (x)$ for $x$ and substitute that in too.
Integrate, then substitute back. Replace $u$ with $g (x)$ and add $+ C$ .

Worked examples

Example 1: A bracket and its derivative

Easy

Integrate:

\int 2 x (x^{2} + 1)^{4} d x

The inside is $x^{2} + 1$ and its derivative, $2 x$ , is sitting right there. Let $u = x^{2} + 1$ , so $d u = 2 x d x$ .

\int 2 x (x^{2} + 1)^{4} d x = \int u^{4} d u = \frac{u ^{5}}{5} + C

Put $x$ back in place of $u$ . Expanding $(x^{2} + 1)^{4}$ and integrating term by term would also work, eventually, after a lot of binomial coefficients. Substitution does it in one line.

Answer

\int 2 x (x^{2} + 1)^{4} d x = \frac{( x ^{2} + 1 ) ^{5}}{5} + C

Example 2: A logarithm in the denominator

Medium

Integrate, for $x > 1$ :

\int \frac{1}{x ln x} d x

This one does not look like it has an inside at all, until you split it as $\frac{1}{l n x} \cdot \frac{1}{x}$ . Then $ln x$ is the inside and $\frac{1}{x}$ is its derivative.

Let $u = ln x$ , so $d u = \frac{1}{x} d x$ :

\int \frac{1}{x ln x} d x = \int \frac{1}{u} d u = ln ∣ u ∣ + C

Substitute back. The absolute value matters: $ln x$ is negative for $x < 1$ .

Answer

\int \frac{1}{x ln x} d x = ln ∣ ln (x) ∣ + C

Example 3: When an x is left over

Harder

Integrate:

\int x x + 1 d x

Let $u = x + 1$ , so $d u = d x$ . That deals with the root, but leaves a stray $x$ outside it. Solve for it: $x = u - 1$ .

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

Now it is two powers, which the power rule handles in reverse: add one to the power, divide by the new power.

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

Swap $u$ back for $x + 1$ . This trick, substituting for the leftover $x$ , is what turns a lot of "substitution doesn't work here" integrals into ones where it does.

Answer

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

Your turn

Three to try, checked by the same machine that grades the game. Type the answer however you like; any correct form counts.

Problem 1

Easy

Integrate:

\int x (x^{2} + 3)^{3} d x

Problem 2

Medium

Integrate:

\int 6 x e^{x^{2} + 1} d x

Problem 3

Harder

Integrate, for $x > \frac{1}{2}$ :

\int \frac{5}{x ln ( 2 x )} d x

Common mistakes

Leaving $d x$ in the integral. $\int u^{4} d x$ is not $\frac{u ^{5}}{5}$ . Swap $d x$ for $d u$ properly, constant factor included.
Losing the constant. In $\int x cos (x^{2}) d x$ , $d u = 2 x d x$ means $x d x = \frac{1}{2} d u$ , so the answer is $\frac{1}{2} sin (x^{2}) + C$ . Forgetting the $\frac{1}{2}$ doubles the answer.
Mixing $x$ and $u$ . If the integral still has an $x$ after substituting, either solve for $x$ in terms of $u$ or try a different $u$ .
Not substituting back. The question was in $x$ , so the answer must be too.
Forgetting to change the limits. On a definite integral, either change the limits to $u$ -values or substitute back before plugging in. Never mix the two.

Practice it now

Five minutes, no account

Substitution integrals come up in a Sprint from difficulty 5 to 7: $x$ times a power, sine, cosine or exponential of a quadratic, and $\frac{1}{x l n ( b x )}$ . The $+ C$ is optional; the game accepts any antiderivative.

Play a Sprint See it on the formula sheet

FAQ

How do you know what to pick for u?: Look for the inside of a composition whose derivative also appears. Good first guesses: the bracket under a power, the argument of $e$ or a trig function, the expression under a square root, or a denominator.
What if du is not exactly what is in the integral?: If it is off by a constant, that is fine: divide it out. For $\int x e^{x^{2}} d x$ , $d u = 2 x d x$ , so $x d x = \frac{1}{2} d u$ and the answer is $\frac{1}{2} e^{x^{2}} + C$ . If it is off by a power of $x$ , it is not fine. $\int e^{x^{2}} d x$ has no $x$ to pair with $d u$ , and in fact it has no elementary antiderivative at all, so no choice of $u$ will rescue it.
Is u-substitution the reverse of the chain rule?: Yes. Differentiate $\frac{( x ^{2} + 1 ) ^{5}}{5}$ with the chain rule and you get back $2 x (x^{2} + 1)^{4}$ . That is also the quickest way to check any substitution answer.
When should you use u-substitution instead of integration by parts?: When one part of the integrand is the derivative of another part's inside. If the two factors are unrelated, like $x$ and $e^{x}$ , substitution cannot help and parts is the tool.
Can you use u-substitution on definite integrals?: Yes. Change the limits as well: if $u = x^{2} + 1$ and $x$ runs from 0 to 1, then $u$ runs from 1 to 2. Then you never have to substitute back.
Where can I find u-substitution practice problems?: Right on this page: "Your turn" has three u-substitution practice problems, easy, medium and harder, checked as you type by the same checker that grades Asymptote, with the answer and the first step if you miss. For more practice against the clock, a five-minute Sprint is free and needs no account.

When to use it

Use substitution when you can see a function inside another function and the derivative of that inside is also in the integrand, up to a constant:

x cos (x^{2})

has

x^{2}

inside and

x

(half its derivative) outside;

\frac{l n x}{x}

has

ln x

and its derivative

\frac{1}{x}

. It is always worth trying before integration by parts, because when it works it is much shorter. If the inside's derivative is nowhere to be found, substitution usually turns the problem into a mess, and that is your sign to try something else.

How to do it, step by step

Choose $u$ . Pick the inside of the most awkward-looking piece: the bracket under a power, the argument of

sin

e^{□}

, the thing under a root, the denominator.

Find $d u$ . Differentiate:

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

. Rearrange so you can swap out whatever is in the integrand, for example

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

Rewrite the whole integral in $u$ . No

x

may survive, and no

d x

. If an

x

is left over, solve

u = g (x)

for

x

and substitute that in too.

Integrate, then substitute back. Replace

u

with

g (x)

and add

+ C

Worked examples

Example 1: A bracket and its derivative

Easy

Integrate:

\int 2 x (x^{2} + 1)^{4} d x

The inside is $x^{2} + 1$ and its derivative, $2 x$ , is sitting right there. Let $u = x^{2} + 1$ , so $d u = 2 x d x$ .

\int 2 x (x^{2} + 1)^{4} d x = \int u^{4} d u = \frac{u ^{5}}{5} + C

Put $x$ back in place of $u$ . Expanding $(x^{2} + 1)^{4}$ and integrating term by term would also work, eventually, after a lot of binomial coefficients. Substitution does it in one line.

Answer

\int 2 x (x^{2} + 1)^{4} d x = \frac{( x ^{2} + 1 ) ^{5}}{5} + C

Example 2: A logarithm in the denominator

Medium

Integrate, for $x > 1$ :

\int \frac{1}{x ln x} d x

This one does not look like it has an inside at all, until you split it as $\frac{1}{l n x} \cdot \frac{1}{x}$ . Then $ln x$ is the inside and $\frac{1}{x}$ is its derivative.

Let $u = ln x$ , so $d u = \frac{1}{x} d x$ :

\int \frac{1}{x ln x} d x = \int \frac{1}{u} d u = ln ∣ u ∣ + C

Substitute back. The absolute value matters: $ln x$ is negative for $x < 1$ .

Answer

\int \frac{1}{x ln x} d x = ln ∣ ln (x) ∣ + C

Example 3: When an x is left over

Harder

Integrate:

\int x x + 1 d x

Let $u = x + 1$ , so $d u = d x$ . That deals with the root, but leaves a stray $x$ outside it. Solve for it: $x = u - 1$ .

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

Now it is two powers, which the power rule handles in reverse: add one to the power, divide by the new power.

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

Swap $u$ back for $x + 1$ . This trick, substituting for the leftover $x$ , is what turns a lot of "substitution doesn't work here" integrals into ones where it does.

Answer

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

Your turn

Three to try, checked by the same machine that grades the game. Type the answer however you like; any correct form counts.

Problem 1

Easy

Integrate:

\int x (x^{2} + 3)^{3} d x

Problem 2

Medium

Integrate:

\int 6 x e^{x^{2} + 1} d x

Problem 3

Harder

Integrate, for $x > \frac{1}{2}$ :

\int \frac{5}{x ln ( 2 x )} d x

Common mistakes

Leaving $d x$ in the integral.

\int u^{4} d x

is not

\frac{u ^{5}}{5}

. Swap

d x

for

d u

properly, constant factor included.

Losing the constant. In

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

d u = 2 x d x

means

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

, so the answer is

\frac{1}{2} sin (x^{2}) + C

. Forgetting the

\frac{1}{2}

doubles the answer.

Mixing $x$ and $u$ . If the integral still has an

x

after substituting, either solve for

x

in terms of

u

or try a different

u

Not substituting back. The question was in

x

, so the answer must be too.

Forgetting to change the limits. On a definite integral, either change the limits to

u

-values or substitute back before plugging in. Never mix the two.

Practice it now

Five minutes, no account

Play a Sprint See it on the formula sheet

FAQ

How do you know what to pick for u?

Look for the inside of a composition whose derivative also appears. Good first guesses: the bracket under a power, the argument of

e

or a trig function, the expression under a square root, or a denominator.

What if du is not exactly what is in the integral?

If it is off by a constant, that is fine: divide it out. For

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

d u = 2 x d x

, so

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

and the answer is

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

. If it is off by a power of

x

, it is not fine.

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

has no

x

to pair with

d u

, and in fact it has no elementary antiderivative at all, so no choice of

u

will rescue it.

Is u-substitution the reverse of the chain rule?

Yes. Differentiate

\frac{( x ^{2} + 1 ) ^{5}}{5}

with the chain rule and you get back

2 x (x^{2} + 1)^{4}

. That is also the quickest way to check any substitution answer.

When should you use u-substitution instead of integration by parts?

When one part of the integrand is the derivative of another part's inside. If the two factors are unrelated, like

x

and

e^{x}

, substitution cannot help and parts is the tool.

Can you use u-substitution on definite integrals?

Yes. Change the limits as well: if

u = x^{2} + 1

and

x

runs from 0 to 1, then

u

runs from 1 to 2. Then you never have to substitute back.

Where can I find u-substitution practice problems?

Right on this page: "Your turn" has three u-substitution practice problems, easy, medium and harder, checked as you type by the same checker that grades Asymptote, with the answer and the first step if you miss. For more practice against the clock, a five-minute Sprint is free and needs no account.

U-Substitution: Steps, Examples & Practice Problems

The formula

When to use it

How to do it, step by step

Worked examples

Example 1: A bracket and its derivative

Example 2: A logarithm in the denominator

Example 3: When an x is left over

Your turn

Common mistakes

FAQ

Keep going

U-Substitution: Steps, Examples & Practice Problems

The formula

When to use it

How to do it, step by step

Worked examples

Example 1: A bracket and its derivative

Example 2: A logarithm in the denominator

Example 3: When an x is left over

Your turn

Common mistakes

FAQ

Keep going