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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.

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In This Article

  1. What Is U-Substitution?

  2. How to Do U-Substitution

  3. Two Common Mistakes Using U-Substitution

  4. Examples of U-Substitution

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 dudu and uu.

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:

ddxf(g(x))=f(g(x))g(x)\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?

f(g(x))g(x)dx\int f(g(x))g’(x)dx

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

f(g(x))g(x)dx=f(u)du\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 uu to find dudu. If your dudu does not match what’s left inside the integrand perfectly, you must rearrange your dudu so that it does match perfectly.

  3. Substitute uu and dudu 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 3(3x+1)5dx\int 3(3x + 1)^5 dx. Examine this function carefully. First, notice that we have a composite of functions with 3x+13x + 1 raised to the fifth power. And, notice that we have a function and its derivative in the same integral. The derivative of 3x+13x + 1 is 33, and both 3x+13x + 1 and 33 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 f(u)du\int f(u)du. Since 3x+13x + 1 is part of a composite of functions, and since the derivative of 3x+13x + 1 is present elsewhere in the integrand, we’ll let u=3x+1u = 3x + 1.

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

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

f(g(x))g(x)dx=f(u)du\int f(g(x))g’(x)dx = \int f(u)du
3(3x+1)5dx=u5du\int 3(3x + 1)^5 dx = \int u^5 du
=u66=\frac{u^6}{6}

4. Finally, we can substitute the original values u=3x+1u = 3x + 1 and du=3dxdu = 3dx back into our equation to get our final answer.

u66=(3x+1)66+C\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 ff and gg are continuous functions.

Sum and Difference Rule:

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

Constant Multiplier Rule:

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

Power Rule:

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

Constant Rule:

adx=ax+C\int adx = ax + C for some constant aa

Reciprocal Rules:

1xdx=x1dx=lnx+C\int \frac{1}{x}dx = \int x^{-1}dx = \ln{|x|} + C

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

Exponential and Logarithmic Function Rules:

exdx=ex+C\int e^xdx = e^x + C

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

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

Trigonometric Function Rules:

For these rules, assume that xx is in radians.

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

cosxdx=sinx+C\int \cos{x}dx = \sin{x} + C

sec2xdx=tanx+C\int \sec ^2 xdx = \tan{x} + C

csc2xdx=cotx+C\int \csc ^2 xdx = -\cot{x} + C

secxtanxdx=secx+C\int \sec{x}\tan{x}dx = \sec{x} + C

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

Two Common Mistakes Using U-Substitution

Choosing the wrong uu or dudu

Many students struggle with determining which expression to designate as uu and which expression to designate as dudu. Choosing the wrong uu and dudu 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 uu 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 uu correctly to find dudu.

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

Sometimes, your calculated dudu isn’t immediately visible inside the integrand. In this case, you can algebraically adjust your dudu 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 xsin(x2+1)dx\int x \sin{(x^2+1)}dx.

Solving for uu and dudu, we find the u=x2+1u = x^2+1 and du=2xdxdu = 2xdx. But it’s clear that du=2xdxdu = 2xdx isn’t part of our integrand, and can’t be immediately substituted.

This is easily fixed. While du=2xdxdu = 2xdx isn’t present, notice that xdxxdx is. We can solve for xdxxdx and substitute that value into our integrand instead. Dividing both sides of du=2xdxdu = 2xdx by 2, we find that 12du=xdx\frac{1}{2}du = xdx.

Now, we can substitute u=x2+1u = x^2+1 and 12du=xdx\frac{1}{2}du = xdx into the integrand and solve normally.

f(g(x))g(x)dx=f(u)du\int f(g(x))g’(x)dx = \int f(u)du
xsin(x2+1)dx=12sin(u)du\int x \sin{(x^2+1)}dx = \frac{1}{2} \int \sin{(u)}du
=12cos(u)= -\frac{1}{2} \cos{(u)}
=12cos(x2+1)+C = -\frac{1}{2} \cos{(x^2+1)} + C

Examples of U-Substitution

We’ll work through 3 more practice problems together.

Example 1

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

Solution:

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

f(g(x))g(x)dx=f(u)du\int f(g(x))g’(x)dx = \int f(u)du
3(3x+1)2dx=duu2\int \frac{3}{(3x+1)^2}dx = \int \frac{du}{u^2}
=u2du= \int u^{-2}du
=u11 = \frac{u^{-1}}{-1}
=13x1+C= \frac{1}{-3x-1} + C

Example 2

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

Solution:

Let u=5x+2u = 5x + 2. Then, du=5dxdu = 5dx. But notice that 5dx5dx is not immediately visible in our integrand. We can fix this by cleverly manipulating the integrand so that 5x5x becomes present. We’ll use the trick that 155=1\frac{1}{5} \cdot 5 = 1. So, we have:

5x+2dx=1555x+2dx\int \sqrt{5x+2}dx = \int \frac{1}{5} \cdot 5 \sqrt{5x+2}dx

Now we can substitute uu and dudu as normal, and use the constant multiplier rule to bring the constant 12\frac{1}{2} outside of the integral.

f(g(x))g(x)dx=f(u)du\int f(g(x))g’(x)dx = \int f(u)du
1555x+2dx=15udu\int \frac{1}{5} \cdot 5 \sqrt{5x+2}dx = \frac{1}{5} \int \sqrt{u}du
=15u12du= \frac{1}{5} \int u^{\frac{1}{2}}du
=15u3232=\frac{1}{5} \cdot \frac{u^{\frac{3}{2}}}{\frac{3}{2}}
=1523u32=\frac{1}{5} \cdot \frac{2}{3}u^{\frac{3}{2}}
=215u32=\frac{2}{15}u^{\frac{3}{2}}
=215(5x+2)32+C=\frac{2}{15}(5x+2)^{\frac{3}{2}} + C

Example 3

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

Solution:

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

f(g(x))g(x)dx=f(u)du\int f(g(x))g’(x)dx = \int f(u)du
ln(x)2xdx=u2du\int \frac{\ln{(x)}^2}{x}dx = \int u^2 du
=u33= \frac{u^3}{3}
=ln(x)33+C= \frac{\ln{(x)}^3}{3} + C

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