  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.

## 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 values 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 values $u = 3x + 1$ and $du = 3dx$ 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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