Integration by Parts: Techniques and Examples

Integration by parts simplifies complex integrals by breaking them into manageable parts.

### Mastering Integration by Parts: Techniques and Examples
* 📘 To find the indefinite integral of \( x e^x \), we apply integration by parts, where \( u = x \) and \( dv = e^x dx \), leading to the result \( **x e^x - e^x + C** \).
* 🔍 Next, for \( x \sin x \), we again set \( u = x \) and \( dv = \sin x dx \), resulting in **\(-x \cos x + \sin x + C\)** after applying the integration by parts formula.
* 📊 When integrating \( x^2 \ln x \), we choose \( u = \ln x \) and \( dv = x^2 dx \), yielding **\(\frac{1}{3} x^3 \ln x - \frac{1}{9} x^3 + C\)** as the final answer.
* ✍ The integral of \( \ln x \) is approached by letting \( u = \ln x \) and \( dv = dx \), resulting in **\( x \ln x - x + C \)**.
* 🔄 For \( x^2 \sin x \), we utilize integration by parts twice, leading to \( **-x^2 \cos x + 2x \sin x + 2 \cos x + C** \).
* 📈 In the case of \( x^2 e^x \), we apply the integration by parts formula twice, yielding \( e^x (x^2 - 2x + 2) + C \) as the solution.
* 📏 The integral of \( \ln(x^2) \) divided by \( x \) is simplified to **\( \frac{1}{3} \ln(x^3) + C \)** through a clever application of logarithmic properties and integration by parts.
* 🎓 Lastly, for \( e^{3x} \cos(4x) \), we perform integration by parts twice, leading to the final answer of **\( \frac{3}{25} e^{3x} \cos(4x) + \frac{4}{3} e^{3x} \sin(4x) + C \)**.

so how can we find the indefinite integral of x times e raised to the x? To do this, we need to use something called integration by parts. The integral of u dv is equal to u times v minus the integral of v du. So, we need to determine u, dv, v, and du. Let me put it in a different order. So, which one should we set equal to u and which one should we set equal to dv? It's best to set u equal to x because if we do so, du will simply be equal to one times dx, and when dealing with constants, they're a lot easier to integrate than variables. So, we're going to make dv equal to e to the x dx. Now, v is the integral of dv. The integral of e to the x is simply e to the x. So now, using the formula u times v, that's going to be x times e raised to the x minus the integral of v du. v is e to the x, du is 1 dx, and the integral of

e to the x is e to the x plus some constant c. So this is the final answer. That's the integral of x raised to the or x times e raised to the x. Let's try another problem. Let's integrate x times sine x dx. So what should we make u and dv equal to? So just like before, I'm going to make u equal to x so that du is going to be 1 dx. Now dv has to be what remains, sine x dx. Now what is the integral of sine? The derivative of what function will give us sine? The integral of sine is negative cosine. The derivative of negative cosine is positive sine. So using the formula, the integral of u dv is equal to u times v minus the integral of v du. So u is x, v is negative cosine x, and then minus v and du is 1 dx. So far we have negative x cosine x, and then these two negative signs will become positive. So plus the integral of cosine x dx. Now the antiderivative of positive

cosine is positive sine, so the final answer is what you see here: negative cosine, I mean negative x cosine x plus sine x plus c. Now here's another problem to work on: let's integrate x squared ln x. So what should we make u and dv equal to in this problem? If we make u equal to x squared and we make dv equal to ln x dx, the integral of ln x will be quite difficult to do unless you know what it is, so we're not going to do that. So it's best to make dv x squared dx and u is going to be ln x because we know the derivative of ln x. The derivative of ln x is 1 over x, and the antiderivative of x squared, the antiderivative of dv, will give us v. So the antiderivative of x squared dx is x cubed over 3. So let's use the formula now: u times v, that's going to be ln x times x cubed over three minus the integral of v times du.

and du is one over x. It's supposed to be dx, so this is equal to one-third x cubed times ln x. And then we could cancel an x, and so we have minus one-third the integral of x squared dx. The integral of x squared is going to be x to the third over three, and then plus some constant c. So the final answer is one-third x cubed ln x minus one over nine x to the third plus c. And that concludes this problem. So let's work on some more examples. Let's try this problem. What is the anti-derivative of the natural log of x? How can we use integration by parts to get this answer? Now we can't make u equal, well, we can't make dv equal to ln x because we don't know the integral of ln x. We're trying to figure that out. So therefore, we have to make u equal to ln x. The derivative of that is one over x dx. dv has to be just, we can make dv dx if we want to.

or one dx and so v the integral of one dx is just x. So now let's start with the formula, and so it's going to be u times v. u is ln x, and v is x, minus the integral of v, and then du is one over x dx. So far, we have x ln x, x times one over x, the x variables cancel, so we have the integral of one dx. Now the integral of one dx is x, so this is the final answer: x ln x minus x plus c. That is the integral of the natural log of x. Let's find the indefinite integral of x squared sine x dx. So which part should be u, and which part is going to be dv? We need to make u equal to x squared, and we need to differentiate it twice to bring it down to a constant. So this is one of those problems where you have to use integration, you know, integration by parts two times. du is going to be 2x dx, dv is sine x dx, so this is dv.

and this is going to be u. The integral of sine is negative cosine. So let's start with the formula. So u times v, that's x squared times negative cosine x, minus the integral of v, which is negative cosine x, times du, that's 2x dx. So let's organize what we have: negative x squared cosine x, and then these two negative signs will become positive, so positive the integral of 2x cosine x. We could take the 2 and move it to the front, so it's just going to be x cosine x. Now we need to use the integration by parts formula on the integral of x cosine x. We're going to make u equal to x, and dv is going to be cosine x dx. The integral of cosine is sine, and the derivative of x is 1 dx. So using the formula, it's going to be u times v, which is x times sine x, minus the integral of v du. So v is sine x, du is just dx, and so the integral of sine is negative cosine.

and then plus c. So now what we need to do is replace this part with what we have here. So it's going to be plus. We need to distribute 2 to everything in here. So it's going to be 2x sine x plus. This would be plus cosine x but times two. So plus two cosine x. And we don't need to multiply c by two because c is just some generic constant. And this whole thing is equal to the original integral. And so this is the answer. Negative x squared cosine x plus 2x sine x plus 2 cosine x plus c. Now let's work on this problem. Let's say we have the integral of x squared e to the x dx. Go ahead and try that problem. So let's begin by writing the formula. The integral of u dv is equal to u times v minus the integral of v du. Now what should we make u equal to and what should be dv? I prefer making dv e to the x dx because once we integrate it, it's going to stay the same.

and I want x squared to eventually become a constant. So, this is one of those problems where you need to use the integration by parts formula twice. So, we're going to make u equal to x squared. du is going to be 2x dx. So, using the formula, it's going to be uv. That's going to be x squared. That's you. v is e to the x minus the integral of v, which is e to the x times du. That's 2x dx. So, now this becomes, well, before we do that, let's use integration by parts one more time. So, I'm going to make u equal to 2x. du is going to be the derivative of 2x, which is 2, but times dx. dv, we're going to make that equal to e x dx, just like we did before, and the integral of that is simply e x. So, we're going to have x squared e x, and then minus now. Let's use the formula again for this part. So, it's uv. So, 2x

e to the x and then minus the integral of vdu, so that's 2 e to the x dx, or multiplying those two. Now let's begin by distributing the negative signs, so this is going to be negative 2x e to the x, and then these two negatives will cancel, giving us positive. Well, let's take out the two, so positive two integral e to the x dx. So our final answer is going to be x squared e to the x minus 2x e to the x. The anti-derivative of e to the x is just e to the x, and then plus c. Now, if we want to, we can factor out e to the x, so we're going to have x squared minus 2x plus 2, and then plus the constant c. So this right here is our final answer. That is the anti-derivative of x squared e to the x. What about this problem? What is the integral of ln x squared dx? Try that. So for this problem, what we need to do is we need to make u.

equal to ln x squared. dv is going to be equal to the other part. We're going to make dv equal to dx, so v is going to be the integral of dx, which is x. d u, I'm going to write that here. I need more space. We need to use the chain rule, so first let's use the power rule by moving the 2 to the front. So it's going to be 2, and then we'll keep everything on the inside the same. 2 times ln x, and then we're going to subtract that by 1, and then take the derivative of the inside. The derivative of ln x is 1/x. So using the formula, this is going to be u times v. So v is x, u is ln x squared, minus the integral of v d u. v is x, d u is 2 ln x times 1/x, and then let's not forget the dx part. So we can cancel x and one of x, so this becomes x ln x squared, and then we could take out the two.

can move it to the front, so minus two integral ln x dx. Earlier in this video, we determined that the integral of ln x is x ln x minus x. So this is going to be -2, and then we can replace this with x ln x minus x. Now let's go ahead and distribute the two, so our final answer is going to be x times ln x squared minus 2x ln x, and then negative 2 times negative x, that's going to be positive 2x, and then plus c. So this is the indefinite integral of ln x squared. Now what would you do if you were to see a problem like this ln x to the seventh power dx? Go ahead and try that. Well, first you need to realize that we can rewrite this problem. A property of logs allows us to move the exponent to the front, so make sure you understand this. This is not ln x to the seventh power like this. If it was, we can't move the seven. The seven is only affecting the x variable. As a result, we can move the 7

to the front, so this becomes 7 ln x. So, this is what we're integrating. This is equivalent to 7 integral ln x dx, and we know the anti-derivative of ln x. It's x ln x minus x. So, this is our final answer. It's just 7 x ln x minus 7 x plus c. Now, let's try this one. Let's find the indefinite integral of e raised to the x sine x dx. Go ahead and work on that. So, for this problem, I want to make u equal to sine x, and then dv, I'm going to set that equal to e to the x dx. So, v is simply going to be e to the x. du is the derivative of sine x, which is cosine x. So, using the formula, it's going to be uv. That's going to be e sine x minus the integral of v du. So, v is e to the x, du is cosine x dx. So, we need to use integration by parts one more time, but this time we're going to make

u equal to cosine x instead of sine, and dv is going to be the same. dv is going to be e to the x dx. If v is the same, du is going to be the derivative of cosine, which is negative sine, and then times dx. So now we're going to have e^x sine x minus, and then in brackets, u v. So that's going to be e^x cosine x minus the integral of v to u. So v is e to the x, du is negative sine x, and then dx. So let's go ahead and distribute the negative sign. I'm going to rewrite the original integral. So the integral of e^x sine x, that's going to be e to the x sine x. These two negative signs will become positive, and then we need to distribute this negative. So it's negative e raised to the x cosine x, and then we need to apply the negative to this, which is going to be negative e to the x sine x. Now the key to solving this problem is

to realize that this term is very similar to the original problem. So, what we want to do is we want to add this to both sides. If we move it to the other side, we're going to have 2 e^x sin x, which is equal to this. Now, we want to find the value of just one integral e^x sin x. So, what we need to do is divide both sides by two. So, we can say that our final answer is one-half. We also factor out e^x, so it's one-half e^x, and then times sin x minus cos x plus c. So, that's the anti-derivative of e raised to the x times sin x. Now, let's move on to our next example. Let's find the antiderivative of ln(x^2) divided by x dx. So, go ahead and try that now. The first thing that I want to do is I'm going to rewrite this as ln(x^2) times 1/x dx. I want to multiply as a product of two

functions. Now what I'm going to do is I'm going to make u equal to ln x squared. du will be 2 times ln x to the first power times 1 over x and then dx. So since I made u ln x squared, dv is going to be the other part. dv is going to be 1 over x dx. So the anti-derivative of 1 over x is ln x. Now using the formula, this is going to be u times v. u is ln x squared, v is just ln x, minus the integral of vdu. So v is ln x, du is 2 ln x times 1 over x dx. So let's put this all together. So first I'm going to rewrite the original problem. That's equal to ln x squared times ln x. That's ln x raised to the third power in this 2. I'm going to move to the front here. We have ln x times ln x. That's ln x squared, and it's multiplied by one over x, which we can move that to the bottom.

and we need to realize that these two terms are similar. So, we need to add this to both sides, or if we move it to the other side, we'll basically add in 1 plus 2, which will become 3. So, on the left side, we're going to have 3 integral ln x squared over x dx, and that's going to equal ln x to the third power. So, if we divide both sides by 3, this 3 will disappear, and we're going to have it underneath here, and then plus our constant c. So, this is the answer to the original problem. That's how we can find the integral of ln x squared over x. Now, for the sake of practice, go ahead and try this: finding a definite integral of e to the 3x cosine 4x dx. So, whenever you have an e to the x times the sine or cosine, it's good to make the cosine part equal to u. That's what I'd like to do personally. So, in this one, I'm going to make u equal to cosine.

4x du. The derivative of cosine is going to be negative sine 4x, and then times the derivative of the angle. The derivative of 4x is 4, and then we're going to have times dx. So dv in this problem is going to be e to the 3x to dx. The antiderivative of e to the 3x is e to the 3x over 3, which we can write as one-third e to the 3x. So using the formula, it's going to be uv, so that's one-third e to the 3x times cosine 4x minus the integral of v du. So v is one-third e to the 3x, du that's negative four sine 4x dx. So what we're going to do now is we're going to move the constants to the front. The two negative signs will cancel, they will become positive, and we can move one-third times four to the front. So this is going to be plus four over three.

integral e to the three x sine four x dx. Now we need to use integration by parts one more time, so dv will be the same. We don't need to adjust this part of the formula, but u is going to be sine 4x as opposed to cosine for x. So du is going to change as well. The derivative of sine is cosine 4x, and then times 4 times dx. So using integration by parts, we're going to have u times v, so that's one-third e to the 3x times sine 4x, and then minus the integral of vdu. So v is one-third e to the three x, and then du, that's times four cosine four x dx. Now let's go ahead and distribute four over three, so multiply those together. That's going to be 4 over 9 e to the 3x sine 4x, and then this 2 will also be 4 over 9, but it's going to be negative four over nine actually. There's a four here, so this is four

thirds times four-thirds, so that's 16 over nine with a negative sign. And then we're going to have the integral of e to the 3x cosine 4x dx. Now notice that these two are similar, so we're going to add 16 over 9 integral e to the 3x cosine 4xdx to both sides. So this is 1, but we can treat it as if it's 9 over 9 just to get common denominators. So 9 over 9 plus 16 over 9, that's going to be 25 over 9 integral e to the 3x cosine 4x dx. And then that's going to be 1/3 e to the 3x cosine 4x, and then plus 4 over 9 e to the 3x sine 4x. Now to get this to 1, what we're going to do is we're going to multiply both sides by the reciprocal of 25 over nine. So that is by nine over twenty-five. So those two fractions will cancel.

on the left side, we're just going to have e to the three x cosine four x dx. And what I'm going to do, I want to keep this in factored form. So we have 9 over twenty-five from one-third and four over nine. We can factor out one-third, and we can also factor out e to the three x. So here, we took out e to the three x, we're left with just cosine four x. Here, we took out e to the three x, and we took out one-third from four over nine. So we're gonna have four-thirds left over times sine 4x. The last thing we could do is maybe reduce this. So if we divide both numbers by 3, this becomes 3 over 25 e to the 3x, and then times cosine 4x plus 4 over 3 sine 4x plus c. So that's going to be the answer for this problem. That's the indefinite integral of e to the 3x cosine 4x.

Integration By Parts

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Integrating Functions Using Integration by Parts

Integration by parts is used to find the indefinite integral of functions like x times e^x.

By setting u = x and dv = e^x dx, we can derive the integral.

The method is also applied to other functions, such as x times sine x and x squared ln x, by strategically choosing u and dv to simplify the integration process.

The integral of the natural log of x is x ln x - x + c.

The indefinite integral of x squared sine x dx requires integration by parts twice, resulting in -x squared cosine x + 2x sine x + 2 cosine x + c.

The integral of x squared e to the x dx also uses integration by parts twice, leading to the expression x squared e to the x - 2x e to the x + 2 e to the x + c.

Finding the Anti-Derivative of x²e^x and ln(x²)

The anti-derivative of x²e^x is found using integration by parts, resulting in the final answer of e^x(x² - 2x + 2) + c.

For the integral of ln(x²)dx, integration by parts is used again, leading to the final answer of xln(x²) - 2xln(x) + 2x + c.

The power rule for logarithms is applied to rewrite ln(x²) before integrating.

The integral of 7 ln x dx is 7x ln x - 7x + c.

For the integral of e^x sin x dx, integration by parts is used twice, leading to the final answer of (1/2)e^x(sin x - cos x) + c.

The integral of ln(x^2)/x dx is approached by rewriting it as ln(x^2) * (1/x) dx.

The integral of ln(x^2)/x is solved using integration by parts, with u = ln(x^2) and dv = 1/x dx.

The solution involves rewriting the terms and combining them, leading to the equation 3 * integral(ln(x^2)/x) = ln(x)^3.

Dividing by 3 gives the final answer, along with a constant C.

The integration by parts is applied twice, first with u as sin(4x) and then with u as cos(4x).

This leads to a common integral of e^(3x)cos(4x) on both sides of the equation.

The final result is (3/25)e^(3x)(cos(4x) + (4/3)sin(4x)) + C.