## 17Calculus - Integration By Substitution

Integration by substitution is the first major integration technique that you will probably learn and it is the one you will use most of the time. In fact, as you learn more advanced techniques, you will still probably use this one also, in addition to the more advanced techniques, even on the same problem. So, this is a critically important technique to learn.

The main idea of this technique is to reduce the complexity of an integral by introducing a new variable to replace a part of the integrand. Then we convert the entire integrand in terms of the new variable. This should reduce the integral to one that we can evaluate using another technique.

Integration by substitution is sometimes called u-substitution since the letter u is often used as the substitution variable.

If you want a complete lecture on integration by substitution, we recommend this video.

### Prof Leonard - Calculus 1 Lecture 4.2: Integration by Substitution [1hr-33mins-57secs]

video by Prof Leonard

Where Integration by Substitution Comes From

The idea of integration of substitution comes from something you already now, the chain rule. Remember, the chain rule for $$y=f(g(x))$$ looks like $$y'=f'(g(x))g'(x)$$. Integration by substitution reverses this by first giving you $$f'(g(x))g'(x)$$ and expecting you to come up with $$f(g(x))$$. This is easier than you might think and it becomes easier as you get some experience. The key is to know how to choose $$g(x)$$. First, let's see how the chain rule becomes an integral.

As we said above, the chain rule for $$y=f(g(x))$$ is $$y'=f'(g(x))g'(x)$$. This can also be written $$dy/dx=f'(g(x))g'(x)$$ or in differential form $$dy=f'(g(x))g'(x)~dx$$. Now when we integrate both sides we have $$y=\int{ f'(g(x))g'(x)dx }$$. The key is to let $$u=g(x)$$ and then $$du=g'(x)dx$$. The integral can now be written $$y=\int{ f'(u)~du } = f(u)+C = f(g(x))+C$$.

So the key to this technique is knowing what to choose for $$g(x)$$. As mentioned above, the letter u is often used as the substitution variable, letting $$u=g(x)$$.

How To Choose u

Let's watch a video clip with a VERY quick example. In this video she gives some really good advice on what to choose for u.

### Krista King Math - Solving Integrals [1min-39secs]

video by Krista King Math

Whew! That video clip went through the example pretty quickly without much explanation. Let's take it a bit slower and explain what is going on. In the video, the presenter evaluated the integral $$\int{x \sqrt{x^2+1} ~dx}$$. Using her ideas, she chose $$u=x^2+1$$. She actually lays out the solution very well but doesn't allow you much time to digest it. So, go back to the video and pause it at the moment she shows her complete solution and make sure you can follow each step.

Knowing what to choose for u is a skill that you get better at with practice. However, there are some specific guidelines to get you started. Let's go through them one-by-one and work some practice problems along the way to improve your skills.

Guideline 1 - Term Under A Power or Root

The video above gives one possible choice for u, the higher power when you have another term with a power one less. From this example, I would extend that idea a bit to include choosing u as whatever is under a radical or a term with a power (like $$(x^2+3)^3 \to u=x^2+3$$).

Let's look at an example to see how to apply this idea.

Evaluate $$\int{ 3x^2 \sqrt{x^3+1} ~dx}$$.

$$\displaystyle{ \frac{2}{3}(x^3+1)^{3/2} +C }$$

Problem Statement: Evaluate $$\int{ 3x^2 \sqrt{x^3+1} ~dx}$$.
Solution: Using the idea from the video above, we choose $$u=x^3+1 \to du=3x^2~dx \to dx=du/(3x^2)$$.
Now we substitute u and dx into the integral.
$$\displaystyle{ \int{ 3x^2 \sqrt{x^3+1} ~dx} = \int{ 3x^2 \sqrt{u} \frac{du}{3x^2} } }$$
Now the $$3x^2$$ term will cancel in the numerator and denominator leaving
$$\int{ u^{1/2}~du }$$ which we evaluate to get
$$\displaystyle{ \frac{u^{3/2}}{3/2} + C = \frac{2u^{3/2}}{3} + C }$$
Now we need to go back to the original variable x using the original substitution $$u=x^3+1$$

$$\displaystyle{ \frac{2}{3}(x^3+1)^{3/2} +C }$$

Notice in the solution to the last example, that at one point we had x's and u's in the integral. We could not evaluate the integral until it had only the one variable u. So had to cancel the $$3x^2$$ term in the numerator and denominator before we could actually integrate.

Okay, now try some on your own. This is one technique that you want to practice a lot until you can do it in your sleep since you will be using it in almost every integral that you evaluate.

Basic Problems

$$\displaystyle{ \int{(3x-5)^{17}dx} }$$

Problem Statement

$$\displaystyle{ \int{(3x-5)^{17}dx} }$$

Solution

### 1010 video

video by Krista King Math

$$\displaystyle{ \int{x\sqrt{x^2+9}~dx} }$$

Problem Statement

$$\displaystyle{ \int{x\sqrt{x^2+9}~dx} }$$

Solution

### 1012 video

video by Krista King Math

$$\displaystyle{ \int{x^2\sqrt{2x^3-4}~dx} }$$

Problem Statement

$$\displaystyle{ \int{x^2\sqrt{2x^3-4}~dx} }$$

Solution

### 1013 video

video by Krista King Math

$$\displaystyle{ \int{2x(x^2+1)^{30}~dx} }$$

Problem Statement

$$\displaystyle{ \int{2x(x^2+1)^{30}~dx} }$$

Solution

### 1017 video

video by PatrickJMT

$$\displaystyle{ \int{2x(x^2+4)^{100}~dx} }$$

Problem Statement

$$\displaystyle{ \int{2x(x^2+4)^{100}~dx} }$$

Solution

### 1021 video

video by PatrickJMT

$$\displaystyle{ \int{y\sqrt[5]{3y^2+1}~dy} }$$

Problem Statement

$$\displaystyle{ \int{y\sqrt[5]{3y^2+1}~dy} }$$

Solution

### 1318 video

video by PatrickJMT

$$\displaystyle{ \int{(12x+8)(3x^2+4x+1)^9~dx} }$$

Problem Statement

$$\displaystyle{ \int{(12x+8)(3x^2+4x+1)^9~dx} }$$

Solution

### 1319 video

video by PatrickJMT

Intermediate Problems

$$\displaystyle{\int{ \frac{\ln x}{x} ~dx} }$$

Problem Statement

$$\displaystyle{\int{ \frac{\ln x}{x} ~dx} }$$

Solution

### 2583 video

$$\displaystyle{ \int{\frac{1}{(4x+7)^6}dx} }$$

Problem Statement

$$\displaystyle{ \int{\frac{1}{(4x+7)^6}dx} }$$

Solution

### 1011 video

video by Krista King Math

$$\displaystyle{ \int{\frac{4}{(3x+1)^5}~dx} }$$

Problem Statement

$$\displaystyle{ \int{\frac{4}{(3x+1)^5}~dx} }$$

Solution

### 1317 video

video by PatrickJMT

$$\displaystyle{ \int{\frac{4x}{\sqrt{x^2+1}}dx} }$$

Problem Statement

$$\displaystyle{ \int{\frac{4x}{\sqrt{x^2+1}}dx} }$$

Solution

### 1016 video

video by Krista King Math

$$\displaystyle{ \int{\frac{(\ln(x))^{10}}{x}dx} }$$

Problem Statement

$$\displaystyle{ \int{\frac{(\ln(x))^{10}}{x}dx} }$$

Solution

### 1014 video

video by Krista King Math

$$\displaystyle{\int{ \csc x \cot x \sqrt{1-\csc x}~dx} }$$

Problem Statement

$$\displaystyle{\int{ \csc x \cot x \sqrt{1-\csc x}~dx} }$$

$$(2/3)(1-\csc x)^{3/2}+C$$

Problem Statement

$$\displaystyle{\int{ \csc x \cot x \sqrt{1-\csc x}~dx} }$$

Solution

### 2106 video

video by PatrickJMT

$$(2/3)(1-\csc x)^{3/2}+C$$

Guideline 2 - Exponential

In the case of an exponential, choose u as the exponent. Since we know how to evaluate $$\int{e^u~du}$$, this choice should significantly simplify the integral.

Basic Problems

$$\displaystyle{ \int{e^{-x}dx} }$$

Problem Statement

$$\displaystyle{ \int{e^{-x}dx} }$$

$$\displaystyle{\int{e^{-x}dx}=-e^{-x}+C}$$

Problem Statement

$$\displaystyle{ \int{e^{-x}dx} }$$

Solution

Evaluate $$\displaystyle{ \int{e^{-x}dx} }$$.
For this integral we need to use the technique of substitution. Since we know that $$\displaystyle{\int{e^{t}dt}=e^t+C}$$, we let $$u=-x\to du=-dx\to-du=dx$$
$$\displaystyle{ \begin{array}{rcl} \int{e^{-x}dx} & = & \int{e^u (-du) } \\ & = & \int{-e^u du} \\ & = & -e^u + C \\ & = & -e^{-x} + C \end{array}}$$

$$\displaystyle{\int{e^{-x}dx}=-e^{-x}+C}$$

$$\displaystyle{ \int{\frac{e^x+1}{e^x}dx} }$$

Problem Statement

$$\displaystyle{ \int{\frac{e^x+1}{e^x}dx} }$$

Solution

### 1018 video

video by PatrickJMT

$$\displaystyle{ \int{\frac{e^{\sqrt{x}}}{\sqrt{x}}dx} }$$

Problem Statement

$$\displaystyle{ \int{\frac{e^{\sqrt{x}}}{\sqrt{x}}dx} }$$

Solution

### 1020 video

video by PatrickJMT

Intermediate Problems

$$\displaystyle{ \int{ (x^2+x)e^{2x^3+3x^2}~dx }}$$

Problem Statement

$$\displaystyle{ \int{ (x^2+x)e^{2x^3+3x^2}~dx }}$$

$$(1/6)e^{2x^3+3x^2} + C$$

Problem Statement

$$\displaystyle{ \int{ (x^2+x)e^{2x^3+3x^2}~dx }}$$

Solution

### 2102 video

video by PatrickJMT

$$(1/6)e^{2x^3+3x^2} + C$$

$$\displaystyle{ \int{e^{x+e^x}~dx} }$$

Problem Statement

$$\displaystyle{ \int{e^{x+e^x}~dx} }$$

Solution

### 1023 video

video by PatrickJMT

Guideline 3 - Trig Functions

When sine and cosine functions are involved, choose u as one of them. When you take the differential, the other one will be part of the differential. For example, $$u=sin(x) \to du=cos(x)~dx$$. Also, it sometimes helps to choose u as the angle of a trig function, if the angle is complicated. For example, if you have $$\sin(3t)$$, let $$u=3t$$.
The same rules apply for hyperbolic trig functions as well.

Basic Problems

$$\displaystyle{ \int{(\sin x) e^{\cos x}~dx} }$$

Problem Statement

$$\displaystyle{ \int{(\sin x) e^{\cos x}~dx} }$$

$$-e^{\cos x} + C$$

Problem Statement

$$\displaystyle{ \int{(\sin x) e^{\cos x}~dx} }$$

Solution

### 2101 video

video by PatrickJMT

$$-e^{\cos x} + C$$

$$\displaystyle{ \int{\tanh(\omega t)~dt} }$$

Problem Statement

$$\displaystyle{ \int{\tanh(\omega t)~dt} }$$

$$\displaystyle{\int{ \tanh(\omega t)~dt}=(1/\omega)\ln(\cosh(\omega t))+C }$$

Problem Statement

$$\displaystyle{ \int{\tanh(\omega t)~dt} }$$

Solution

 $$\displaystyle{\int{ \tanh(\omega t) ~dt } = \int{ \frac{\sinh(\omega t)}{\cosh(\omega t)} ~dt }}$$ $$u = \cosh(\omega t) \to du = \omega \sinh(\omega t) ~dt$$ $$\displaystyle{ \int{\frac{1}{u} \frac{du}{\omega}} }$$ $$\displaystyle{ \frac{1}{\omega} \ln\abs{u} + C }$$ $$\displaystyle{ \frac{1}{\omega} \ln \abs{ \cosh(\omega t) } + C }$$

Since $$\cosh(\omega t) > 0$$ for all $$t$$, we can drop the absolute value signs and write $$\ln \abs{ \cosh(\omega t) }$$ as $$\ln ( \cosh(\omega t) )$$. However, leaving the absolute values signs is also correct.

$$\displaystyle{\int{ \tanh(\omega t)~dt}=(1/\omega)\ln(\cosh(\omega t))+C }$$

Intermediate Problems

$$\displaystyle{ \int{ \left[ \sec(\cos x)\tan(\cos x )\right] \sin x ~dx } }$$

Problem Statement

$$\displaystyle{ \int{ \left[ \sec(\cos x)\tan(\cos x )\right] \sin x ~dx } }$$

$$-\sec(\cos x) + C$$

Problem Statement

$$\displaystyle{ \int{ \left[ \sec(\cos x)\tan(\cos x )\right] \sin x ~dx } }$$

Solution

### 2105 video

video by PatrickJMT

$$-\sec(\cos x) + C$$

Guideline 4 - Fraction Denominator

When it makes sense, sometimes choosing u as the denominator of a fraction is helpful since $$\int{(1/x)dx} = \ln(x)$$. Let's look at an example.

Evaluate $$\displaystyle{ \int{ \frac{x^3}{5+x^4} ~dx} }$$.

$$\displaystyle{ \frac{1}{4}\ln(5+x^4) + C }$$

Problem Statement: Evaluate $$\displaystyle{ \int{ \frac{x^3}{5+x^4} ~dx} }$$.
Solution: Using one of the substitution suggestions above, set u to the denominator, i.e. $$u=5+x^4 \to du=4x^3dx$$.
$$\displaystyle{ \int{ \frac{x^3}{5+x^4} ~dx} = \int{ \frac{x^3}{u}~\frac{du}{4x^3} } }$$
Now, the $$x^3$$ terms cancel and we have an extra 4 in the denominator which we can pull outside the integral to get $$\displaystyle{ \frac{1}{4} \int{ \frac{1}{u}~du } = \frac{1}{4} \ln(u) + C }$$
Substituting back into x terms gives us the final answer.

$$\displaystyle{ \frac{1}{4}\ln(5+x^4) + C }$$

In the last example, notice that all the x terms canceled but we had an extra constant, which we factored out of the integral. This will happen quite often and it helps to move all the constants outside the integral like we did so that it is easy to see that all the variables in the integral are now u's. Once we have that, we can do the actual integration.

Basic Problems

$$\displaystyle{ \int{\frac{5x}{5+2x^2}dx} }$$

Problem Statement

$$\displaystyle{ \int{\frac{5x}{5+2x^2}dx} }$$

Solution

### 1015 video

video by Krista King Math

$$\displaystyle{ \int{\frac{2^x}{2^x+1}dx} }$$

Problem Statement

$$\displaystyle{ \int{\frac{2^x}{2^x+1}dx} }$$

Solution

### 1019 video

video by PatrickJMT

$$\displaystyle{\int{ \frac{x}{x^2+4}~dx } }$$

Problem Statement

$$\displaystyle{\int{ \frac{x}{x^2+4}~dx } }$$

$$(1/2)\ln\abs{x^2+4} + C$$

Problem Statement

$$\displaystyle{\int{ \frac{x}{x^2+4}~dx } }$$

Solution

### 2103 video

video by PatrickJMT

$$(1/2)\ln\abs{x^2+4} + C$$

$$\displaystyle{ \int{ \frac{24x^3-4}{3x^4-2x+1}~dx } }$$

Problem Statement

$$\displaystyle{ \int{ \frac{24x^3-4}{3x^4-2x+1}~dx } }$$

$$2\ln\abs{3x^4-2x+1} + C$$

Problem Statement

$$\displaystyle{ \int{ \frac{24x^3-4}{3x^4-2x+1}~dx } }$$

Solution

### 2104 video

video by PatrickJMT

$$2\ln\abs{3x^4-2x+1} + C$$

Intermediate Problems

$$\displaystyle{ \int{\frac{1}{x^2+4x+13}dx} }$$

Problem Statement

$$\displaystyle{ \int{\frac{1}{x^2+4x+13}dx} }$$

Solution

### 1846 video

video by Dr Chris Tisdell

$$\displaystyle{ \int{ \frac{dx}{1+x^{1/4}} } }$$

Problem Statement

$$\displaystyle{ \int{ \frac{dx}{1+x^{1/4}} } }$$

Solution

### 1009 video

video by PatrickJMT

$$\displaystyle{ \int{ \frac{x}{1-x^2+\sqrt{1-x^2}}dx } }$$

Problem Statement

$$\displaystyle{ \int{ \frac{x}{1-x^2+\sqrt{1-x^2}}dx } }$$

Solution

### 1024 video

video by PatrickJMT

$$\displaystyle{ \int{ \frac{x}{x^2+6x+10}~dx } }$$

Problem Statement

$$\displaystyle{ \int{ \frac{x}{x^2+6x+10}~dx } }$$

Solution

### 1841 video

video by Dr Chris Tisdell

Extra x Terms

Up until now, we have done the substitution of u and all the x terms have canceled. However, what do you do when they do not all cancel? Let's look at an example to see how to do this.

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

$$\displaystyle{ \frac{2}{5}(x+2)^{5/2} -\frac{4}{3}(x+2)^{3/2}+C }$$

Problem Statement: Evaluate $$\int{ x \sqrt{x+2} ~dx }$$.
Solution: Let $$u=x+2 \to du=1~dx$$.
$$\int{ x \sqrt{x+2} ~dx = \int{ x \sqrt{u} ~du } }$$
Notice that we are not able to cancel out the extra x in the integral and we can't do the integration until we have only u's. So what we do is go back to the initial equation for u, which is $$u=x+2$$ and we solve for x giving us $$x=u-2$$. We use this to substitute back into the integral for x to get rid of it. This gives us
$$\int{ (u-2)\sqrt{u} ~du }$$, which we can now evaluate.
$$\int{ (u-2)\sqrt{u} ~du } = \int{ u^{3/2} - 2u^{1/2} ~du } =$$ $$\displaystyle{ \frac{u^{5/2}}{5/2} - 2\frac{u^{3/2}}{3/2}+C }$$
Now we substitute back in for u and we are done.
Final Answer: $$\displaystyle{ \frac{2}{5}(x+2)^{5/2} -\frac{4}{3}(x+2)^{3/2}+C }$$
Note: If we are asked to simplify or factor, our answer will be $$\displaystyle{ 2(x+2)^{3/2}\left[ (x+2)/5 -2/3 \right] +C }$$

$$\displaystyle{ \frac{2}{5}(x+2)^{5/2} -\frac{4}{3}(x+2)^{3/2}+C }$$

Here are some practice problems involving this technique.

$$\displaystyle{ \int{ \frac{x}{\sqrt[4]{x+2}}~dx } }$$

Problem Statement

$$\displaystyle{ \int{ \frac{x}{\sqrt[4]{x+2}}~dx } }$$

Solution

### 1022 video

video by PatrickJMT

$$\displaystyle{ \int{x\sqrt[3]{x+4}~dx} }$$

Problem Statement

$$\displaystyle{ \int{x\sqrt[3]{x+4}~dx} }$$

Hint

For his substitution, he used $$u=\sqrt[3]{x+4}$$. That is a little more difficult than we would recommend. Try letting $$u=x+4$$ and see if this is easier to evaluate.

Problem Statement

$$\displaystyle{ \int{x\sqrt[3]{x+4}~dx} }$$

Hint

For his substitution, he used $$u=\sqrt[3]{x+4}$$. That is a little more difficult than we would recommend. Try letting $$u=x+4$$ and see if this is easier to evaluate.

Solution

### 1025 video

video by PatrickJMT

You CAN Ace Calculus

 basics of integrals differentials substitution from precalculus

### Calculus Topics Listed Alphabetically

Single Variable Calculus

Multi-Variable Calculus

### Search Practice Problems

Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.

free ideas to save on books

 The 17Calculus and 17Precalculus iOS and Android apps are no longer available for download. If you are still using a previously downloaded app, your app will be available until the end of 2020, after which the information may no longer be available. However, do not despair. All the information (and more) is now available on 17calculus.com for free.

When using the material on this site, check with your instructor to see what they require. Their requirements come first, so make sure your notation and work follow their specifications.

DISCLAIMER - 17Calculus owners and contributors are not responsible for how the material, videos, practice problems, exams, links or anything on this site are used or how they affect the grades or projects of any individual or organization. We have worked, to the best of our ability, to ensure accurate and correct information on each page and solutions to practice problems and exams. However, we do not guarantee 100% accuracy. It is each individual's responsibility to verify correctness and to determine what different instructors and organizations expect. How each person chooses to use the material on this site is up to that person as well as the responsibility for how it impacts grades, projects and understanding of calculus, math or any other subject. In short, use this site wisely by questioning and verifying everything. If you see something that is incorrect, contact us right away so that we can correct it.