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Calculus

# What Is U-Substitution?

01.29.2022 • 6 min read

## Rachel McLean

Subject Matter Expert

U-substitution is a useful method for integrating composite functions. In this article, we’ll discuss the meaning of u-substitution, how to use u-substitution, common mistakes students make, and review some practice problems.

In This Article

## What Is U-Substitution

You’re probably familiar with the idea that integration is the reverse process of differentiation. U-substitution is an integration technique that specifically reverses the chain rule for differentiation. Because of this, it’s common to refer to u-substitution as the reverse chain rule. We may also refer to it as integration by substitution, or “change of variables” integration.

U-substitution integration allows us to find the antiderivative of composite functions.

Dr. Tim Chartier explains antiderivatives more:

You can think of u-substitution as the chain rule executed backward. To “undo” the chain rule, we rewrite the integral in terms of $du$ and $u$.

Since u-substitution “undoes” the chain rule, we can use the chain rule formula to help determine which problems require u-substitution. If you can spot a function and its derivative in the same integrand, that indicates that u-substitution is likely the best integration method for that scenario.

## How to Do U-Substitution

First, let’s review the chain rule below. The chain rule states:

$\frac{d}{dx}f(g(x)) = f’(g(x))g’(x)$

An integrand that you can evaluate using u-substitution might look something like this. Do you see the chain rule formula?

$\int f(g(x))g’(x)dx$

Using u-substitution, we substitute $u = g(x)$ and $du = g’(x)$ into our integrand, which “undoes” the chain rule:

$\int f(g(x))g’(x)dx = \int f(u)du$

### 4 Steps to Follow

Here are 4 simple steps for u-substitution:

1. Pick your “u”. This expression is the "inside" part of the chain rule and is usually the term inside a radical, power, or denominator.

2. Differentiate $u$ to find $du$. If your $du$ does not match what’s left inside the integrand perfectly, you must rearrange your $du$ so that it does match perfectly.

3. Substitute $u$ and $du$ into the integrand, and integrate using the key integration formulas that are already familiar to you.

4. Substitute the original value for $u$ back into the equation after integrating. Remember to add the constant of integration to your final answer.

### Practice Problem

Let’s solve one problem together. We’ll evaluate $\int 3(3x + 1)^5 dx$. Examine this function carefully. First, notice that we have a composite of functions with $3x + 1$ raised to the fifth power. And, notice that we have a function and its derivative in the same integral. The derivative of $3x + 1$ is $3$, and both $3x + 1$ and $3$ are inside the integrand. This means that u-substitution is the way to go! Let’s walk through our 4 steps.

1. Remember that we want our substitution integral to look like $\int f(u)du$. Since $3x + 1$ is part of a composite of functions, and since the derivative of $3x + 1$ is present elsewhere in the integrand, we’ll let $u = 3x + 1$.

2. To find $du$, we can find $\frac{du}{dx}$ and then multiply by $dx$. Since $\frac{du}{dx} = 3$, then $du = 3dx$.

3. Now we can evaluate our integral by substituting $u = 3x + 1$ and $du = 3dx$ into the integrand. Then, we can evaluate using the power rule for integrals.

$\int f(g(x))g’(x)dx = \int f(u)du$
$\int 3(3x + 1)^5 dx = \int u^5 du$
$=\frac{u^6}{6}$

4. Finally, we can substitute the original value for $u$, $u = 3x + 1$, back into our equation to get our final answer.

$\frac{u^6}{6} = \frac{(3x+1)^6}{6} + C$

To help with Step 3, review the list of standard integral rules below. These formulas are essential for any integration method. Assume that $f$ and $g$ are continuous functions.

#### Sum and Difference Rule:

$\int [f(x) \pm g(x)]dx = \int f(x)dx \pm \int g(x)dx$

#### Constant Multiplier Rule:

$\int kf(x)dx = k\int f(x)dx$ for some constant $k$

#### Power Rule:

$\int x^ndx = \frac{x^{n+1}}{n+1} + C$ for some real number $n$

#### Constant Rule:

$\int adx = ax + C$ for some constant $a$

#### Reciprocal Rules:

$\int \frac{1}{x}dx = \int x^{-1}dx = \ln{|x|} + C$

$\int\frac{1}{ax+b}dx = \frac{1}{a} \ln{(ax+b)} + C$

#### Exponential and Logarithmic Function Rules:

$\int e^xdx = e^x + C$

$\int a^xdx = \frac{a^x}{\ln{(x)}} + C$, for any positive real number $a$

$\int \ln{(x)}dx = x\ln{(x)}-x + C$

#### Trigonometric Function Rules:

For these rules, assume that $x$ is in radians.

$\int \sin{x}dx = -\cos{x} + C$

$\int \cos{x}dx = \sin{x} + C$

$\int \sec ^2 xdx = \tan{x} + C$

$\int \csc ^2 xdx = -\cot{x} + C$

$\int \sec{x}\tan{x}dx = \sec{x} + C$

$\int \csc{x}\cot{x}dx = -\csc{x} + C$

## Two Common Mistakes Using U-Substitution

Choosing the wrong $u$ or $du$

Many students struggle with determining which expression to designate as $u$ and which expression to designate as $du$. Choosing the wrong $u$ and $du$ will result in an incorrect answer.

Remember, you’re looking for two functions within the integrand that fit the framework given by the chain rule. Make sure that $u$ is equal to the “inside” function of the chain rule, or the inner part of the composite of functions. Double-check that you’ve differentiated $u$ correctly to find $du$.

Forgetting to multiply/divide by a constant, when necessary.

Sometimes, your calculated $du$ isn’t immediately visible inside the integrand. In this case, you can algebraically adjust your $du$ substitution to match the expression that is present in the integrand. Usually, this means multiplying or dividing the integrand by a constant. For example, consider $\int x \sin{(x^2+1)}dx$.

Solving for $u$ and $du$, we find the $u = x^2+1$ and $du = 2xdx$. But it’s clear that $du = 2xdx$ isn’t part of our integrand, and can’t be immediately substituted.

This is easily fixed. While $du = 2xdx$ isn’t present, notice that $xdx$ is. We can solve for $xdx$ and substitute that value into our integrand instead. Dividing both sides of $du = 2xdx$ by 2, we find that $\frac{1}{2}du = xdx$.

Now, we can substitute $u = x^2+1$ and $\frac{1}{2}du = xdx$ into the integrand and solve normally.

$\int f(g(x))g’(x)dx = \int f(u)du$
$\int x \sin{(x^2+1)}dx = \frac{1}{2} \int \sin{(u)}du$
$= -\frac{1}{2} \cos{(u)}$
$= -\frac{1}{2} \cos{(x^2+1)} + C$

## Examples of U-Substitution

We’ll work through 3 more practice problems together.

### Example 1

Evaluate $\int \frac{3}{(3x+1)^2}dx$.

### Solution:

Let $u = 3x + 1$. Then, $du = 3dx$. Now, we can substitute $u$ and $du$ into the integrand.

$\int f(g(x))g’(x)dx = \int f(u)du$
$\int \frac{3}{(3x+1)^2}dx = \int \frac{du}{u^2}$
$= \int u^{-2}du$
$= \frac{u^{-1}}{-1}$
$= \frac{1}{-3x-1} + C$

### Example 2

Evaluate $\int \sqrt{5x+2}dx$.

### Solution:

Let $u = 5x + 2$. Then, $du = 5dx$. But notice that $5dx$ is not immediately visible in our integrand. We can fix this by cleverly manipulating the integrand so that $5x$ becomes present. We’ll use the trick that $\frac{1}{5} \cdot 5 = 1$. So, we have:

$\int \sqrt{5x+2}dx = \int \frac{1}{5} \cdot 5 \sqrt{5x+2}dx$

Now we can substitute $u$ and $du$ as normal, and use the constant multiplier rule to bring the constant $\frac{1}{2}$ outside of the integral.

$\int f(g(x))g’(x)dx = \int f(u)du$
$\int \frac{1}{5} \cdot 5 \sqrt{5x+2}dx = \frac{1}{5} \int \sqrt{u}du$
$= \frac{1}{5} \int u^{\frac{1}{2}}du$
$=\frac{1}{5} \cdot \frac{u^{\frac{3}{2}}}{\frac{3}{2}}$
$=\frac{1}{5} \cdot \frac{2}{3}u^{\frac{3}{2}}$
$=\frac{2}{15}u^{\frac{3}{2}}$
$=\frac{2}{15}(5x+2)^{\frac{3}{2}} + C$

### Example 3

Evaluate $\int \frac{\ln{(x)}^2}{x}dx$.

### Solution:

Let $u = \ln{(x)}$. Then, $du = \frac{1}{x}dx$. Now, we can substitute $u$ and $du$ into the integrand.

$\int f(g(x))g’(x)dx = \int f(u)du$
$\int \frac{\ln{(x)}^2}{x}dx = \int u^2 du$
$= \frac{u^3}{3}$
$= \frac{\ln{(x)}^3}{3} + C$

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