Substitution for Indefinite Integrals: Videos & Practice Problems

To evaluate integrals involving composite functions, the substitution method, or u-substitution, is essential. By selecting an appropriate u, typically the inner function, and finding du, the integral can be simplified. For example, in the integral of $x^2+\; 1$ cubed times $2x$ dx, choosing $u=x^2+\; 1$ simplifies the integral to $u^3$ du. This method streamlines the integration process, allowing for easier evaluation and application of the power rule.

Indefinite Integrals

Since we've been working through evaluating various integrals, let's take a look at another integral here. The integral of x2 + 1 cubed times 2xdx. Observing this integral, we can see that we have a function inside of another function, which is then being multiplied by yet another function. Currently, we do not have the tools needed to evaluate this integral as presented. We could proceed by expanding x2 + 1 three times and distributing the 2x throughout, but that would be quite a tedious process.

Luckily, there's a much easier way to evaluate this integral. From earlier calculus studies, we learned to evaluate the derivatives of composite functions using the chain rule. Now, faced with evaluating integrals that include composite functions, we'll use a process called substitution, also referred to as u-substitution or change of variables. Integrals are the reverse process of finding derivatives and, likewise, substitution is essentially the reverse process of using the chain rule.

I'm going to walk you through this substitution process here step by step. So, let’s reconsider our integral of x2 + 1 cubed times 2xdx. We identified that we have the function x2 + 1 inside another function, since it's being cubed. Examining this integral in its entirety reveals a specific form, where a composite function f ( g ( x ) ) is multiplied by g'(x).

Given an integral of this form, we can use substitution to simplify evaluation, allowing us to rewrite it as f ( u ) du, where u is a placeholder variable used for our substitution. This appears significantly easier to evaluate than the original integral. The critical step is determining what u and du should be. When making a substitution, u will often be our inside function, situated inside a parenthesis, under a radical, or in an exponent.

Let's analyze our example and choose u. We see that x2 + 1 is enclosed in parentheses and is being cubed. Given this arrangement inside the parentheses, this indicates a good candidate for u.

If u=x2 + 1, then what is du? Taking the derivative with respect to x, du/dx=2x, we then move the dx to the other side to give du = 2xdx, exactly what our integral includes. This allows the substitution of the integral to simplify to u3du.

Evaluating this new integral is straightforward using the power rule. Adding one to the exponent and dividing by the new exponent gives 14u4+c, where c is the constant of integration, as this is an indefinite integral. Having successfully performed our substitution and derived our integral, we return it to terms of x since u=x2+1. We can then rewrite this as 14(x2+1)4+c, finalizing our original integral evaluation using the substitution process.

In this example, we observed that when we chose u=x2+1, du neatly appeared in our integral. However, this scenario won't always occur, which is why breaking it down step-by-step equips you to deal with any integral. We echoed these steps in solving the above example, now articulated. For a new task, the integral to evaluate is integral of 4 4x-1dx.

Let’s initiate using the same steps by identifying u and finding du. In this scenario, 4x-1 under the radical suggests selecting it as u. Consequently, du = d(4x-1)=4dx. This completes step one.

Moving to step two, we rewrite the integral solely in terms of u and du. Here, u appears under the radical, and we have a solo dx. Since du includes 4dx, not just dx, there's an offset by a constant. By multiplying by four, we achieve our necessary du, but to preserve the integral's value, we simultaneously multiply by one-fourth, effectively canceling the four.

Thus, u as 4x-1 and 4dx=du lets us express the integral fully in terms of u and du: 14×integraludu. With step two finalized, we proceed to the third.

Integrating with respect to u, we know u equates to u12, allowing us to utilize the power rule. Retaining the constant of one-fourth, we perform 14×23u32+c, simplifying to 16u32+c,

In our final step, we restore the original variable x by substituting back u with 4x-1. Now, our answer to the integral is 16(4x-1)32+c. Addressing other integrals in similar fashion aids in mastery, and further substitution practice is up next. I’ll see you there.

Indefinite Integrals Example 1

In this problem, we are asked to use substitution in order to evaluate the given integral. Then, after we do that, we want to then check our answer by differentiating. So, let's go ahead and get started here by using substitution to evaluate this integral, the indefinite integral of 2x3+x+75 times 6x2+1 dx. Now, using substitution here, we're just going to jump right into our steps. Our very first step is going to be to choose u and find du.

Now, remember, u is going to be an inside function, typically in parentheses, under a radical, in an exponent, or something like that. Now, in this case, we can see that there are two functions that are enclosed within parentheses, so you may be confused as to which one should be u. Now we can see that one of these functions is raised to a power whereas the other one is not. So that's usually a good clue that this is the inside function that we should choose as u. So let's go ahead and see how that works out.

Choosing u to be that 2x3+x+7. Then once we have u, we want to find du, which is going to be the derivative of u times dx. So, the derivative of this 2x3+x+7 using the power rule is going to be 6x2+1, and then we multiply that by dx. So, we can already see based on our integral that this is going to work out rather nicely. So, with step one done, we can go ahead and rewrite our integral only in terms of this u and du.

So, I can see that if I chose this as a u, that then made this du, that 6x2+1 dx. So, because of that, we can rewrite this integral as a u5 du. Now, with step two done, we can move on to step three where we just want to go ahead and integrate with respect to u. Now here, we can do that rather easily using the power rule here, adding one to that exponent and then dividing by that new exponent. That gives me a 16u6 and, of course, plus my constant of integration c because this is an indefinite integral.

So, with that step three done, we move on to our final step in finding this integral, which is replacing u with that original function of x. Now, remember, we chose u to be that 2x3+x+7, so that's what we want to replace u as here to get our final answer. This is going to give us a 16 times that 2x3+x+76 plus my constant of integration at c. So, this is my final answer here, having evaluated that integral using a substitution. Now remember in this specific example, we were told that we want to go ahead and check our answer by differentiating our result here.

So, if we did this correctly, when we take the derivative of this function here, we should get the original function that was in our integrand. So, let's go ahead and find this derivative, ddx of 162x3+x+76 plus that constant of integration c. Now, looking at this, I can clearly see that we need to use the chain rule when finding this derivative. So, I'm going to go ahead and start by identifying that this is that inside function. So, when I take the derivative, I'm first going to start on the outside and work my way in.

So, this 16 times everything in parentheses to the power of six. I'm going to go ahead and use the power rule here. We want to multiply by our original exponent and then decrease that exponent by one. So, taking this derivative, I'm going to get six times 16, keeping that original constant, times 2x3+x+7, keeping that inside function the same for now. And then this is now raised to the power of five having taken the derivative.

Now, applying the chain rule, we then want to multiply by the derivative of that inside function. Now, the derivative of 2x3+x+7 using the power rule is going to give me 6x2+1. So, if I go ahead and do some canceling here, six and 16 cancel. So, this is going to be 2x3+x+75 and then times 6x2+1. So, we can see that the final result that we got here from differentiating the answer that we got when taking the integral is exactly what we started with in our integrand, which means that we did this correctly.

Now, if you have any questions, feel free to let us know in the comments, and I'll see you in the next one.

Indefinite Integrals Example 2

Here we're asked to evaluate the indefinite integral of x2eex3 dx. Now we're going to do this using substitution. Let's go ahead and get started with our first step, choosing u and finding du. Remember, when we choose u, this is most often going to be some inside function either in parentheses, under a radical, or in an exponent. Here, we can see that eis raised to the power of x3. So that means that x3 is going to be a good choice for u.

So if I choose u to be x3, what does that make du? Remember du is going to be the derivative of that x3 times dx. Here that is du = 3 x2 dx. Now I can see in my integral I have that x2 but I don't have that three. So, as we move into step two here, rewriting our integral only in terms of u and du, we need to go ahead and take care of that by multiplying by a constant and its reciprocal in order to get the du that we want to make our substitution work.

I'm going to go ahead and rewrite the order that this is in just to make sure we can clearly see what's going on here. If I have this eex3, this is then multiplied by x2dx. Now I want this to be 3x2dx. So I can go ahead and multiply by that three. But when I do that, I also need to multiply by the reciprocal of three, so I'm effectively canceling it out.

So if I multiply by the reciprocal of three, that's just one-third. Then when I'm working with this integral, I can pull that one-third out to the front of my integral so that it's just a constant and I have everything I need in my integral right here. So if this is u, then that makes this du. Rewriting my integral only in terms of u and du, this is going to be one third times the integral of eeu du. Moving into step three here, integrating with respect to u, I know that the antiderivative of eex is just eex.

The same thing is true for eeu. So this then becomes a one-third times eeu plus the constant of integration c. Then I need to go ahead and take my final step here which is replacing u with that original function of x. Remember, that was x3 so that makes this equal to one-third eex3 plus that constant of integration c, and this is my final answer. If you have any questions, let us know and let's keep practicing.

Evaluate the indefinite integral.
$\int_{}^{}3t\sqrt{t^2+7}dt$

• A. $t^{\frac{3}{2}}+C$
• B. ${(t^2+7)}^{\frac{2}{3}}+C$
• C. ${(t^2+7)}^{\frac{3}{2}}+C$
• D. $t^{\frac{2}{3}}+C$

Answer: (C)

Evaluate the indefinite integral.
$\int_{}^{}\frac{1}{\left(3x+2\right)^5}dx$

• A. $-\frac{1}{12{(3x+2)}^4}+C$
• B. $-\frac{1}{5{(3x+2)}^4}+C$
• C. $\frac{1}{12{(3x+2)}^4}+C$
• D. $\frac{1}{5{(3x+2)}^4}+C$

Answer: (A)

Evaluate the indefinite integral.
${\int }_{}^{}\frac{1}{2x+5}dx$

• A. $\frac{1}{2}\ln x+C$
• B. $\frac{1}{2}\ln {(2x+5)}^2+C$
• C. $2\ln (2x+5)+C$
• D. $\frac{1}{2}\ln (2x+5)+C$

Answer: (D)

Substitution With an Extra Variable

Earlier, we learned how to evaluate an integral using the process of substitution. And when making a substitution, we saw that du didn't always just appear nicely in our integral, and it could be off by some factor of a constant. And as it turns out, our integral may also be off by some extra variable, like some extra factor of x or x2 or x plus or minus a constant, and we need to account for that when making our substitution. And that's exactly what I'm going to show you how to do here by just adding one extra step within those steps that we've already learned. So let's go ahead and get started here by jumping right into our example where we're asked to evaluate the integral by making a substitution.

And here we have the integral of x×x+3 dx. So getting started with our steps the same way we have before, we want to go ahead and choose u and find du. Now looking at my integral here, since I have this x+3 underneath a radical, that is a good indicator that that's what I should choose to be u. And if u is equal to x+3, that then means that du is just equal to dx because the derivative of x+3 is just one. So, with that U and DU that I found in my integral here, we can go ahead and move on to step two, which is rewriting our integral only in terms of that U and DU.

Now based on u and du here, if I wanted to go ahead and rewrite my integral, I would have the integral of x×u du. Now looking at this, this is clearly not only in terms of u and du because we have this extra factor of x here. So how can we get rid of this x and put this whole integral just in terms of u and du? Well, since I know what u is as a function of x, it's just equal to x+3, I can go ahead and just rearrange this equation. So instead of having u in terms of x, I can get x in terms of u.

So if I go ahead and just subtract three from both sides here, I am then left with x being equal to u-3. And if x is equal to u-3, I can go ahead and just replace that in my integral. So rewriting this integral, this is then the integral of u-3 times the square root of u du. And now this is only in terms of u and du, just like we wanted. So we can now proceed to step three, solving this as usual.

Now remember that step three is to integrate here with respect to u, and I'm going to go ahead and clean up this integral a littlebit to make it easier to apply our power rule. Remember that the square root of u is the same thing as u12. I'm going to go ahead and distribute that into my parentheses here. So this is going to be the integral of a u1 times u12 adding those exponents together one plus one half that gives me u32 and then minus three times u12} integrated of course with respect to u. Now applying the power rule here to get my antiderivative u32, I'm going to add one to that exponent, which gives me 52.

If I divide by 52, that's the same thing as multiplying by two fifths. So this is two fifths u52 and then minus three, keeping that constant on the outside there, just paying attention to this u to the power of 12. Again, using the power rule, adding one to 12 gives me 32. If I divide by 32, that's the same thing as multiplying by two thirds. So that's two thirds

Evaluate the indefinite integral.
$\int_{}^{}x\left(5+x\right)^{79}dx$

• A. $\frac{1}{81}{(5+x)}^{81}-\frac{1}{16}{(5+x)}^{80}+C$
• B. $\frac{1}{81}{x}_{}^{81}-\frac{1}{16}{x}^{80}+C$
• C. $\frac{1}{81}{(5+x)}^{81}-\frac{1}{80}{(5+x)}^{80}+C$
• D. $\frac{1}{81}{x}^{81}-\frac{1}{80}{x}^{80}+C$

Answer: (A)

Evaluate the indefinite integral.
$\int_{}^{}\frac{t}{\sqrt{t-2}}dt$

• A. $\frac{2}{3}{t}^{\frac{3}{2}}+4t^{\frac{1}{2}}+C$
• B. $\frac{2}{3}{(t-2)}^{\frac{2}{3}}+2+C$
• C. $\frac{2}{3}{(t+2)}^{\frac{3}{2}}+4{(t+2)}^{\frac{1}{2}}+C$
• D. $\frac{2}{3}{(t-2)}^{\frac{3}{2}}+4{(t-2)}^{\frac{1}{2}}+C$

Answer: (D)

Evaluate the indefinite integral.
$\int_{}^{}\frac{x}{\left(x-6\right)^5}dx$

• A. $-\frac{x}{6{(x-6)}^6}-\frac{3x}{6{(x-6)}^6}+C$
• B. $-\frac{1}{3{(x-6)}^6}-\frac{3}{2{(x-6)}^6}+C$
• C. $-\frac{1}{3{(x-6)}^3}-\frac{3}{2{(x-6)}^4}+C$
• D. $\frac{1}{3{(x-6)}^3}+\frac{3}{2{(x-6)}^4}+C$

Answer: (C)

Substitution With an Extra Variable Example 3

In this problem, we're told that the marginal revenue of a coffee truck selling x lattes is given by the function below. \( R'(x) \) is equal to \( \frac{2x}{\sqrt{3x^2 - 50}} \). Now the first thing that we're asked to do here is to find the total revenue function \( R(x) \) if the revenue from selling 15 lattes is $75. Now since we're asked here to find \( R(x) \) and we're given \( R'(x) \), that means that we need to integrate this \( R'(x) \) in order to get our total revenue function \( R(x) \). So let's start there.

\( R(x) \) is going to be equal to the integral of \( \frac{2x}{\sqrt{3x^2 - 50}} \, dx \). So looking at this, I can tell that we need to do a substitution because I have this function underneath my radical. So in order to perform this substitution, I'm going to set \( u \) equal to what's under the radical here, which is \( 3x^2 - 50 \). Now if \( u \) is equal to \( 3x^2 - 50 \), what does that make \( du \)? Well, \( du \) is going to be equal to the derivative of that \( 3x^2 - 50 \), which is \( 6x \, dx \).

So with that \( 6x \, dx \) and that \( 3x^2 - 50 \), I can see in my integral here that this is \( u \), but then I have \( 2x \, dx \) in my integral, which means that I need to go ahead and multiply by a constant and its reciprocal in order to make this substitution work. So how can I get from \( 2x \, dx \) to \( 6x \, dx \) that I want? Well, I need to multiply by three. If I multiply this by three, I also need to multiply my integral by one-third here in order to cancel that out. But now three times two x is \( 6x \), and this is exactly what \( du \) is.

So I can rewrite this integral as \( \frac{1}{3} \), the integral of \( \frac{1}{\sqrt{u}} \, du \). Now \( \frac{1}{\sqrt{u}} \), I can also just write this as a power, \( u^{-\frac{1}{2}} \), so that I can more easily apply the power rule here when finding my antiderivative. So looking at this here, this is going to be \( \frac{1}{3} \times \) (using the power rule here, I'm going to add one to that exponent and then I'm dividing by that new exponent, that \( -\frac{1}{2} + 1 \). \( -\frac{1}{2} + 1 \) is positive \( \frac{1}{2} \).

If I'm dividing by positive \( \frac{1}{2} \), that means that I'm effectively multiplying by two. So this is two \( u^{\frac{1}{2}} \), and then plus my constant of integration \( c \). Now we can go ahead and combine our two constants here. \( \frac{1}{3} \times 2 \) is \( \frac{2}{3} \). So this is \( \frac{2}{3} \times u \), which I can go ahead and plug back in.

It's that \( 3x^2 - 50 \), and it's to the power of one half, which just means that it's under a square root. So \( 3x^2 - 50 \) and then plus that constant of integration \( c \). So this is my function for my total revenue, \( R(x) \), but we were given some additional information here. We were told that the revenue from selling 15 lattes is $75. So that means that we can actually figure out what this constant of integration is so that we can actually have an accurate total revenue function.

So in order to do that, we need to plug in 15 for \( x \) because that's what \( x \) represents, my number of lattes, and then $75 is my revenue \( R \). So this means that 75 is equal to \( \frac{2}{3} \) times the square root of \( 3 \times 15^2 - 50 \) plus that constant of integration \( c \). Now since this constant of integration is sitting over here just being added, that means that in order to get that constant of integration by itself and ultimately solve for it, I just need to subtract all of this over to the other side. So let's go ahead and do that here. So if I subtract this from 75, this will mean that \( c \) is then equal to 75 - \( \frac{2}{3} \) times the square root of \( 3 \times 15^2 - 50 \) and I can go ahead and plug all of this in my calculator all at once making sure to use parentheses where appropriate, and I will ultimately find that \( c \) is equal to 58.33, going ahead and rounding that to two decimal places.

So now that I know what my constant of integration is, I can fully write my total revenue function correctly with that constant plugged in here. So with this constant now that we know \( R(x) \) is equal to \( \frac{2}{3} \times \sqrt{3x^2 - 50} + 58.33 \), that constant of integration that we found based on our initial conditions. So we have our revenue function here. And the next question that we're asked is how many lattes must be sold for our truck to have a revenue of at least $250? So we want all of this over here to equal 250.

So I want to plug in 250 for \( R \) and then solve for \( x \). So I have that $250 is equal to \( \frac{2}{3} \times \sqrt{3x^2 - 50} + 58.33 \). So given this equation, we need to solve it for \( x \), which just means that we have a lot of algebra to do in order to get that \( x \) isolated all by itself. So the first thing that we need to do here is go ahead and subtract this 58.33 from both sides. When we do that, we will see it cancel over here on the right and then 250 minus that 58.33 is going to give me here 191.67.

So that 191.67, just rewriting this, 191.67 is equal to \( \frac{2}{3} \times \sqrt{3x^2 - 50} \) and we canceled out that constant on the side. Now the next thing that we want to do here is get rid of this \( \frac{2}{3} \). So how can we get rid of \( \frac{2}{3} \)? Well, we can multiply by \( \frac{3}{2} \). That will effectively cancel that fraction out.

Whatever I do to one side, I have to do to the other side. So that means I'm multiplying this 191.67 by that \( \frac{3}{2} \) as well. We will see this cancel over here on the right yet again. And then on this left side, we are left with 287.505. I'm avoiding rounding for now, because I want my answer to be as accurate as possible.

So that 287.505 is equal to what's left over here on this right side, which is \( 3x^2 - 50 \). So what is our next algebra step here? We're trying to get that \( x \) by itself. So our next step is to get rid of this square root by squaring. Whatever I do to one side, I have to do to the other side, so I'm squaring that left side as well.

We'll see this radical cancel over here on the right, and then 287.505 squared is going to give us a rather large number. It's 82,659.12503. Again, I'm avoiding rounding until the very end here. So this is equal to what's left under my radical. That's \( 3x^2 - 50 \).

So my next step here, we're left with a much easier equation now that we've gotten to this point. We need to add 50 to both sides. Now when I add 50 to both sides here, that will cancel out this negative 50. And then over here on the left adding 50 to this, this is going to be 82,709.12503 and that's equal to \( 3x^2 \). So next step here, divide both sides by three to get that three to cancel out.

And we are finally left here with 27,569.70834, and that's equal to \( x^2 \). So one final step here in order to get \( x \) all by itself, we need to go ahead and square root both sides. So if I square root this side and square root this side, this is going to cancel over here and then I am left with \( x \) being equal to the square root of this big long number. When we plug that into our calculator here, we will ultimately get that \( x \) is equal to 166.04, going ahead and rounding that off to two decimal places. Now typically when we're working with a square root, we want to take plus or minus our value.

But because we're working with lattes being sold, we can't sell a negative number of lattes, so we only took our positive result here. So we need to sell 166.04 lattes, but can we sell 0.04 of a latte? Well, no. We definitely can't do that. So that means that we need to go ahead and round this up to the nearest whole number, because if I only sell 166 lattes, that's not enough to get me to that $250 revenue.

But if I sell 167 lattes, then I'm going to be a little bit over that $250, but I'll have at least gotten there. So, ultimately, we want to sell 167 lattes in order to have a revenue of at least $250. Now, feel free to let us know if you have any questions here, and I'll see you in the next one.

Substitution With an Extra Variable Example 4

If you haven't already given this problem a try on your own, go ahead and do that before checking back in with me. In this example, we're told to answer the following questions based on the population growth rate given to us below, where p of t is a population of a particular species of fish present in a lake after t years. Now the population growth rate that we're given here is p′t=350⁢et3. Again, t is t years. So we want to find here p of t given that there are 5,000 fish in the lake at t equals zero, so when zero years have passed.

So because we want to find p of t here, the population, we want to go ahead and integrate our growth rate. So, to find p of t, we need to take the integral of 350 times et3 dt. Now, to find this integral, we need to perform a substitution because I can see I have this t over three in that exponent here. If I make u equal to t over three, that then makes du equal to one-third dt. In order to get that one-third dt, I need to go ahead and multiply by one-third in my integral here.

And if I multiply by one-third, I also need to multiply it by the reciprocal of one-third, which is three. And I'm going to put that on the outside of my integral because I don't need it inside to make that substitution. Alright. Let's go ahead and keep going here. I know that this is du and then this t over three is u.

I'm going to go ahead and pull all my constants out front so that three and that 350. Three times 350 is 1,050, so I'm just going to stick that right out front there. And then this is 1,050 times eu du. Now I know that the antiderivative of e to the u is just e to the u, so this is 1,050 e to the u plus that constant of integration c. Of course, we want to go ahead and replace that u with what we know it to be t over three.

So that makes this function 1,050 et3 plus that constant of integration c for this population pt. But in our problem statement, we were told there were 5,000 fish present in the lake when t is equal to zero. So we can use that information to find this constant. We want to plug in zero for t and 5,000 for p. So this is 5,000 is equal to 1,050 times e0 over three, plugging that zero in for t, plus c.

Now e0 is just one. So this term goes to one and one thousand and fifty times one is just 1,050. So all we need to do here is subtract this 1,050 to get c by itself. Whenever we do that, we'll then find that c is equal to 3,950. Now that we know what our constant is, we can plug it back into our function in order to actually get what p of t is.

So here this population function is going to be equal to 1,050 et3 and then plus that constant that we found, 3,950. So this is my population function here. Now the next thing we're asked is to find the population of fish after six years, and we want to round that to the nearest whole number. So that means that we just need to plug six for t into our population function that we just found in part a. So, in order to find p of six, this is 1,050 times e63 plus 3,950. Now we can just type all of this into our calculator all at once. And whenever we do that, we should get an answer here of 11,708.5089. We were told to round to the nearest whole number, so going ahead and rounding this gives us 11,709, rounding that up because of that decimal. So 11,709 fish present after six years. Now feel free to let us know if you have any questions here, and I'll see you in the next one.

Here’s what students ask on this topic:

What is the substitution method for indefinite integrals?

The substitution method for indefinite integrals is a technique used to simplify the integration of composite functions. It involves choosing a new variable, $u$, which is typically the inner function of the composite function. The differential, $du$, is then determined. This substitution transforms the integral into a simpler form, often allowing the use of the power rule to find the antiderivative. After integrating, the result is expressed back in terms of the original variable. This method is particularly useful for integrals that resemble the derivative of a composite function, making it easier to evaluate complex integrals efficiently.

How do you choose $u$ in substitution for integrals?

Choosing $u$ in substitution for integrals involves identifying the inner function of a composite function. Typically, $u$ is selected as the expression inside parentheses, under a radical, or in an exponent. The goal is to simplify the integral by transforming it into a function of $u$ and $du$. Once $u$ is chosen, its derivative is calculated to find $du$. This choice should make the integral easier to evaluate, often allowing the use of the power rule. It's important to ensure that the substitution accounts for all parts of the original integral, including any constants or extra variables.

What is the role of the power rule in substitution for indefinite integrals?

The power rule plays a crucial role in substitution for indefinite integrals by providing a straightforward method to find antiderivatives. Once the integral is rewritten in terms of $u$ and $du$, the power rule can be applied to integrate functions of the form $u^n$. The power rule states that the antiderivative of $u^n$ is $\frac{u}{\mathrm{n+1}}$ divided by $\mathrm{n+1}$, plus a constant of integration. This simplifies the integration process, allowing for efficient evaluation of complex integrals. After applying the power rule, the result is converted back to the original variable to complete the integration.

How do you handle constants in substitution for indefinite integrals?

Handling constants in substitution for indefinite integrals involves adjusting the integral to match the differential $du$. If $du$ is off by a constant factor, you can multiply the integral by that constant and its reciprocal to maintain the integral's value. For example, if $du$ is $4dx$ but the integral has $dx$, multiply by $4$ and $\frac{1}{4}$. This ensures the substitution works without altering the integral's value. This technique is essential for correctly applying substitution and simplifying the integration process.

What are the steps to perform substitution in indefinite integrals?

The steps to perform substitution in indefinite integrals are as follows: 1) Choose $u$, the inner function of the composite function, and find $du$ by differentiating $u$. 2) Rewrite the integral in terms of $u$ and $du$. If necessary, adjust for constants by multiplying by the constant and its reciprocal. 3) Integrate the new expression using the power rule or other appropriate methods. 4) Substitute back the original variable to express the final result in terms of the original variable. These steps simplify the integration of complex functions, making it more manageable.