Finding the integral of can be tricky at first glance. I’m sure you’re probably familiar with how to take the integral of the part of it. In general, you’ll just want to remember that $$\int e^x \ dx = e^x$$

However, the piece looks a little weird. At first glance it may even look completely foreign. What have you seen that looks like this piece of the function?

Let’s think about what’s going on here

If we look back at our original integral, you’ll want to notice the dx at the end of it. Remember we had $$\int e^x + x^e \ dx$$

The dx is important because it indicates what letter is our variable that we will be integrating with respect to. Since it’s dx, that means we will be integrating with respect to x. Therefore, e will need to be treated as a constant.

In fact, e is a known constant with a specific value. It’s kind of like . We know that . So when we are trying to integrate the part of this function, what we are integrating just comes down to “a variable raised up to a constant power.”

When you are trying to integrate x raised up to some constant power, you would want to use the power rule. The power rule for integrating tells us that if n is some constant other than n = -1, then $$\int{x^n} \ dx \ = \ \frac{x^{n+1}}{n+1} \ = \ \frac{1}{n+1} \cdot x^{n+1}$$

So since e is a constant, we can basically just replace the n in the above formula with e to find the integral of . Using this, we can see that $$\int{x^e} \ dx \ = \ \frac{x^{e+1}}{e+1} \ = \ \frac{1}{e+1} \cdot x^{e+1}$$

Putting these integrals together

Now that we have figured out how to integrate the part of our function, let’s go back to the original integral. As we do this, let’s also think about one of the basic integral properties. Specifically, the one that tells us what to do about integrating a function that is a sum of two simpler functions. $$\int f(x)+g(x) \ dx \ = \int f(x) \ dx + \int g(x) \ dx$$

So as a result of this, we can break down our problem into two simpler integrals that we already know. $$\int e^x + x^e \ dx \ = \int e^x \ dx + \int x^e \ dx$$

And since we already found both of these integrals earlier, we know $$\int e^x + x^e \ dx \ = \ e^x + \frac{x^{e+1}}{e+1} + C$$

U-substitution is one of the more common methods of integration. It allows us to find the anti-derivative of fairly complex functions that simpler tricks wouldn’t help us with. The best way to think of u-substitution is that its job is to undo the chain rule.

That’s all we’re really doing.

It’s not too complicated when you think of it that way. Although, the execution isn’t always that simple. But I’ll show you 6 simple steps that will help you solve any u-substitution problem!

1. Picking our u

A u-substitution problem will start out similarly to an integration by parts problem. With any u-substitution problem the first thing you will need to do is decide what piece of the function you will call u. This is the most important piece of the process, and really the only part where there are options to choose from. However, there’s a simple trick to make sure you’re selecting the u correctly.

When deciding which part of your function to call u, you will want to look for a piece of your function that you can see that piece’s derivative somewhere else in the function. That sounds a little strange, but let me give you an example.

Say we have some function like . We want to look for a small piece of this function that also has its own derivative somewhere else in the function.

So we might say because the derivative of is just 2x. Notice, our function contains an x, not a 2x, but it’s fine if the derivative differs by a constant like this. It’s easy to deal with the constant here, but it’s important that it’s an x term.

A quick note on substitution

Choosing the correct u in these problems is the most challenging part. It’s not always simple to see what the u should be, so it’s important to be willing to try different things and see what happens.

You may end up needing to pick a u to go a few steps into the problem and realize it won’t work, then go back and pick another u. I know this process can be frustrating at times, we’ve all been there, but sometimes trial and error is required in learning new math concepts. So I urge you to stay persistent and keep picking different parts of the function for u.

If you try calling every possible piece of the function u, and work through the next few steps only to find that none of them will work, you likely need to either use some uncommon trick or find the derivative using another method besides u-substitution. I will get into some of the uncommon tricks in a later post.

2. Finding du

Once you have decided which piece of your function will be u, you then need to calculate du. This should be fairly simple.

All you have to do to find du is take the derivative of u then multiply it all by dx. This will sometimes require the use of the chain rule, product rule, or quotient rule, but usually you will just need the power rule.

Let’s think back to our previous example, finding the antiderivative of . Remember, we decided that we would use . Therefore, to find du we should take it’s derivative and multiply by dx. This means if

$$u=x^2+5, \ then$$

$$du=2x \ dx.$$

3. Solve for dx

Now we have determined which part of our function we will call u and we found du. However, what we really want is to find dx. This will be easier to work with when we do our substitution into the original function.

All we need to do to find dx is take our equation for du and isolate the dx. In this example, this will be very easy.

$$du=2x \ dx$$

$$\frac{du}{2x}=dx$$

4. Substitute back into the original function

Going back to the function we are trying to find the antiderivative of, we will first write this in integral form.

$$\int x(x^2+5)^3 dx$$

Now we just need to substitute our u and dx back into this integral. This requires replacing the with u, and replacing dx with . This gives us

$$\int x(u)^3 \frac{du}{2x}$$

After making these two substitutions we should be able to do some simplifying that will cancel out any remaining x‘s. Simplifying this integral should leave us with

$$\int \frac{1}{2}u^3 du.$$

And since we can pull constants out of an integral, this can also be written as

$$\frac{1}{2} \int u^3 du.$$

5. Integrate with respect to u

Looking at the above integral, we can see that we no longer have an x in the problem. We have rewritten everything in terms of u. Since our integral contains only u and du, instead of x and dx, we can integrate with respect to u. This simply means that we are taking the antiderivative of the function where u is our variable.

Notice, this integral is much simpler than the one we started with. All we need to find this one is the power rule for antiderivatives.

$$\frac{1}{2} \int u^3 du$$

$$\frac{1}{2} \cdot \frac{1}{4}u^4$$

$$\frac{1}{8} u^4$$

Now the hard part is over, we found the antiderivative. But unfortunately we aren’t quite done yet. We can’t say that is the antiderivative of . This doesn’t really have any meaning because they are using two different variables. Instead we need to write the answer in terms of x, since that’s what we started with.

6. Substitute x back in

We are almost done now that we found the antiderivative. We just need to write it in terms of x so that our answer actually is the antiderivative of the function we started with. Since we already found the antiderivative in terms of u, and we know u in terms of x, we can simply substitute in for u.

We decided back in step 1 that

$$u=x^2+5$$

and we also found out that our antiderivative in terms of u is

$$\frac{1}{8}u^4.$$

Therefore, we can plug in for u to find that the antiderivative of is

$$F(x)= \frac{1}{8}(x^2+5)^4 + c.$$

And that’s it! You can apply these 6 steps to solve any u-substitution problem.

I hope this lesson helps, but if there’s still a topic you’d like to learn about take a look at some of my other lessons and problem solutions. If you canāt find the topic or question youāre looking for just let me know by emailing me at jakesmathlessons@gmail.com!

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