Name: Integration By Parts
Uploaded: 2025-05-05T15:42:55.000Z
Duration: 32 min 52 s
Description: Thousands of YouTube videos with English-Chinese subtitles! Now you can learn to understand native speakers, expand your vocabulary, and improve your pronunciation...

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.

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 1 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.

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.

That's the integral of x times e raised to the x.

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.

The derivative of what function will give a sine?

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 sines 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.

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 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 u times v, that's going to be ln x times x cubed over 3 minus the integral of v times du, and du is 1 over x.

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 3, and then plus some constant c.

So the final answer is one-third x cubed ln x minus 1 over 9 x to the third plus c, and that concludes this problem.

What is the antiderivative 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.

So therefore, we have to make u equal to ln x.

The derivative of that is 1 over x dx. dv has to be just, we can make dv dx if we want to, or 1 dx, and so v, the integral of 1 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 1 over x dx.

So far we have x ln x, x times 1 over x, the x variables cancel, so we have the integral of 1 dx.

Now, the integral of 1 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 of, you know, integration of our 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.

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.

Negative x squared cosine x, and then these two negative signs will become positive, so positive sine x dx.

We can take the two 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 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 2, so plus 2 cosine x.

And we don't need to multiply c by 2, 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.

Let's say we have the integral of x squared e to the x dx.

I would 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 u. v is e to the x minus the integral of v, which is e to the x times du.

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 to the x dx, just like we did before.

And the integral of that is simply e to the x.

So we're going to have x squared, e to the x, and then minus.

Now let's use the formula again for this part.

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Integration By Parts

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