u-substitution

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 - use this when one part of the function is inside another, and its derivative (or something close to the derivative) is also present. By temporarily using a new variable and simplifying the integral, integration rules can be applied more easily.

Steps for $u$ -substitution

Choose $u = g (x)$ .

Pick the inner function - look for something nested or a function whose derivative also appears elsewhere. Don’t use $u = x$ , since that’s just a letter change.

Differentiate $u$

Take the derivative

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

and rearrange so that

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

Rewrite the integral

Substitute $u$ and $d u$ into the integral. There should be no $x$ anywhere.

Integrate with respect to $u$ .

Use the integration rules from the previous section.

Plug $g (x)$ back in.

If it’s an indefinite integral, don’t forget the $+ C$ .

Examples

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

Solution

There’s an inner function $x^{2}$ nested in the cosine, so let’s try

$u = x^{2}$

Then

$d u = 2 x d x$

which appears in the integral so rewriting the integral 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 adjust by multiplying or dividing by a constant.

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

Solution

(spoiler)

There’s a nested $x^{2}$ again. Let’s try

$u = x^{2}$

Then

$d u = 2 x d x$

However, since only $x d x$ appears in the integrand, let’s adjust by dividing by $2$ , so that

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

Then rewriting the integral in terms of $u$ ,

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

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

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

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

AP tip:

For quotients, a general recommendation is to set $u$ as the bottom expression or a nested portion of it. Otherwise there’s no way to reach a $g^{'} (x)$ that’s in the denominator.

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

Solution

(spoiler)

Let’s try

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

Then

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

Notice that $2$ can be factored out such that

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

And since $(x - 1) d x$ appears in the integrand, then

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

Rewriting the integral in terms of $u$ ,

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

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

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

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

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

Solution

(spoiler)

This involves a quotient but setting $u = x$ in the denominator won’t get us anywhere. However, because the derivative of $ln (x)$ is $\frac{1}{x}$ which does appear in the integrand, let’s set $u$ to be the entire nested function

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

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

Then the integral rewritten in terms of $u$ is

$2 \int u d u$

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

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

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

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

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

Solution

(spoiler)

There’s a nested function $2 x$ , but recall that the derivative of $tan (x)$ is $sec^{2} (x)$ . Since a function and its derivative appear in the integrand, it would be better to choose

$u = tan (2 x)$

then

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

Rearranging,

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

The integral rewritten is

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

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

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

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

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

The reason both are correct is because of the Pythagorean identity $tan^{2} (x) + 1 = sec^{2} (x)$ , where the two functions 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 an interesting one that involves applying an additional technique during the $u$ -substitution.

$\int x x - 1 d x$

Solution

(spoiler)

There’s a nested function $x - 1$ , so let’s try

$u = x - 1$

Then

$d u = d x$

But how do we take care of the $x$ outside the square root?

What we can do is rearrange the chosen $u$ by adding $1$ to both sides so that

$u + 1 = x$

Then the integral rewritten in terms of $u$ is

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

Distribute $u$ to get

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

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

Then using the reverse power rule, the integral is

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

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

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

Solution

(spoiler)

Let

$u = x^{2} + 1$

$d u = 2 x d x$

Notice that in the numerator, $x^{3}$ can be broken apart into $x^{2} \cdot x$ . Rearrange the chosen $u = x^{2} + 1$ to get

$x^{2} = u - 1$

Rearrange $d u$ to get

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

Which means

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

Then the integral rewritten in terms of $u$ is

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

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

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

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

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

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

Definite integrals and changing bounds

If you’re using $u$ -substitution on a definite integral, there are two options to evaluate it using the FTC part 2:

Option 1: Change the bounds

Change the limits from $x$ to $u$ so that the antiderivative does not need to be changed back to a function in terms of $x$ .

Option 2: Don’t change the bounds.

Plug $g (x)$ back into $u$ and evaluate the result with the original bounds.

Examples

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

(spoiler)

Calculations are often a little easier if the constant multiple is brought out first.

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

$(x^{2} + 1)^{2}$ is the denominator, but if we let $u = (x^{2} + 1)^{2}$ , then $d u = 2 (x^{2} + 1) (2 x) d x$ which is more complicated than what appears in the numerator. Instead, try

$u = x^{2} + 1$

Then

$d u = 2 x d x$

In order to get $x d x$ , divide by $2$ .

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

At this point, we have two options:

Option 1: Change the bounds

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

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

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

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

Then the integral rewritten in terms of $u$ is

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

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

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

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

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

$= \frac{6}{5}$

Option 2: Don’t change the bounds

Use the same substitution $u = x^{2} + 1$ to rewrite the integral in terms of $u$ , but keep the original bounds of $x = 0$ to $x = 2$ :

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

After integrating to this step, change $u$ back to $x^{2} + 1$ and evaluate from $0$ to $2$

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

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

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

$= \frac{6}{5}$

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 a function whose derivative also appears elsewhere. Don’t use $u = x$ , since that’s just a letter change.

Differentiate $u$

Take the derivative

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

and rearrange so that

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

Rewrite the integral

Substitute $u$ and $d u$ into the integral. There should be no $x$ anywhere.

Integrate with respect to $u$ .

Use the integration rules from the previous section.

Plug $g (x)$ back in.

If it’s an indefinite integral, don’t forget the $+ C$ .

Examples

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

Solution

There’s an inner function $x^{2}$ nested in the cosine, so let’s try

$u = x^{2}$

Then

$d u = 2 x d x$

which appears in the integral so rewriting the integral 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 adjust by multiplying or dividing by a constant.

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

Solution

(spoiler)

There’s a nested $x^{2}$ again. Let’s try

$u = x^{2}$

Then

$d u = 2 x d x$

However, since only $x d x$ appears in the integrand, let’s adjust by dividing by $2$ , so that

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

Then rewriting the integral in terms of $u$ ,

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

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

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

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

AP tip:

For quotients, a general recommendation is to set $u$ as the bottom expression or a nested portion of it. Otherwise there’s no way to reach a $g^{'} (x)$ that’s in the denominator.

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

Solution

(spoiler)

Let’s try

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

Then

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

Notice that $2$ can be factored out such that

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

And since $(x - 1) d x$ appears in the integrand, then

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

Rewriting the integral in terms of $u$ ,

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

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

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

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

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

Solution

(spoiler)

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

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

Then the integral rewritten in terms of $u$ is

$2 \int u d u$

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

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

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

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

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

Solution

(spoiler)

There’s a nested function $2 x$ , but recall that the derivative of $tan (x)$ is $sec^{2} (x)$ . Since a function and its derivative appear in the integrand, it would be better to choose

$u = tan (2 x)$

then

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

Rearranging,

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

The integral rewritten is

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

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

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

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

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

The reason both are correct is because of the Pythagorean identity $tan^{2} (x) + 1 = sec^{2} (x)$ , where the two functions 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 an interesting one that involves applying an additional technique during the $u$ -substitution.

$\int x x - 1 d x$

Solution

(spoiler)

There’s a nested function $x - 1$ , so let’s try

$u = x - 1$

Then

$d u = d x$

But how do we take care of the $x$ outside the square root?

What we can do is rearrange the chosen $u$ by adding $1$ to both sides so that

$u + 1 = x$

Then the integral rewritten in terms of $u$ is

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

Distribute $u$ to get

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

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

Then using the reverse power rule, the integral is

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

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

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

Solution

(spoiler)

Let

$u = x^{2} + 1$

$d u = 2 x d x$

Notice that in the numerator, $x^{3}$ can be broken apart into $x^{2} \cdot x$ . Rearrange the chosen $u = x^{2} + 1$ to get

$x^{2} = u - 1$

Rearrange $d u$ to get

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

Which means

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

Then the integral rewritten in terms of $u$ is

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

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

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

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

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

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

Definite integrals and changing bounds

If you’re using $u$ -substitution on a definite integral, there are two options to evaluate it using the FTC part 2:

Option 1: Change the bounds

Change the limits from $x$ to $u$ so that the antiderivative does not need to be changed back to a function in terms of $x$ .

Option 2: Don’t change the bounds.

Plug $g (x)$ back into $u$ and evaluate the result with the original bounds.

Examples

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

(spoiler)

Calculations are often a little easier if the constant multiple is brought out first.

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

$(x^{2} + 1)^{2}$ is the denominator, but if we let $u = (x^{2} + 1)^{2}$ , then $d u = 2 (x^{2} + 1) (2 x) d x$ which is more complicated than what appears in the numerator. Instead, try

$u = x^{2} + 1$

Then

$d u = 2 x d x$

In order to get $x d x$ , divide by $2$ .

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

At this point, we have two options:

Option 1: Change the bounds

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

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

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

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

Then the integral rewritten in terms of $u$ is

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

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

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

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

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

$= \frac{6}{5}$

Option 2: Don’t change the bounds

Use the same substitution $u = x^{2} + 1$ to rewrite the integral in terms of $u$ , but keep the original bounds of $x = 0$ to $x = 2$ :

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

After integrating to this step, change $u$ back to $x^{2} + 1$ and evaluate from $0$ to $2$

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

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

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

$= \frac{6}{5}$

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