Find the integral.∫x2lnxdx\int_{}^{}x^2\ln xdx

x 3 3 ln ⁡ x − x 3 9 + C \frac{x^3}{3}\ln x-\frac{x^3}{9}+C

Find the integral. ∫ x 2 ln x d x \int_{}^{}x^2\ln xdx

this problem, we need to find the integral of x squared times the natural log of x. Now how could I solve a problem like this? Well, typically, when we see the product of two functions like this, a first thought is usually to use some kind of variable substitution. Now you could try a variable substitution on this x squared here, but that's actually not gonna get us to a solution since the other x we have here is inside the natural log. If you tried a variable substitution on this x here, then its derivative would just go to one, which would not help us at all with this x squared out in front. So this is gonna be a case where we need to use integration by parts to solve the problem. Now we should be familiar with this formula that the integral of u d v is equal to u v minus the integral of v d u. And when using this formula for integration by parts, our first step is typically going to be whatever this u is. Now u needs to become more simple as we take its derivative. Now this is a problem where things can get a bit confusing, because it may be tempting to take a look at this x squared here and say, well, x squared is just going to become two x when we take its derivative, so that should be what we set equal to u. But here's the thing. Which is gonna become more simple, the x squared or the ln of x? Well, Well, the ln of x, if we take its derivative, would become one over x. That is more simple than than the natural log, and so because of that, I actually think it's gonna be better to set the natural log of x equal to u instead. Now, alternatively, you could use this memory tool that we've talked about previously. The memory tool, lieate, tells us the general order that we need to solve these types of cases where we need to find u. Now the first thing listed here is logarithm. So it goes logarithm, and then we go inverse, then algebraic, then trig, and then exponential. So logarithm would be the first thing that we look for. So this is a case where the memory tool can be really helpful, but for here, we can just generally understand that u is going to be this thing that becomes that becomes more simple when we take the derivative. And in this case, the natural log would become much more simple. Now what I would do from here is set this x squared here equal to d v because dv needs to be something that's easily integrated. Now in this case we can relatively easily integrate these these terms where we have x raised to a power because we know that there's just a power where we can apply. So this is the case where I want to set the dv equal to x squared. So this is one of those cases where you actually don't want to set the algebraic term to the u. Now now if you accidentally did that where you set the algebraic where you set the, the x squared equal to u, you could run through this problem, but you would find you'd end up getting stuck as you went got to this portion where you had to solve the integral. But if we solve it like this, this is going to allow us to get to an answer we can solve. So go ahead and take the derivative of my natural log of x, which I know is just gonna be one over x with respect to x. Now go ahead and integrate d v. Well, that's gonna give me v, which can be the integral of x squared with respect to x. And doing this, well, the integral of x squared is just gonna be x cubed over three using the power rule, and now we've gone ahead and solved for these variables right here. So now that we've found our u and our dv, and we've also found our du and v, I can just plug everything into this side of the equal sign, so let's go ahead and do that. So what this is going to do is equal u times v, which I can see is x cubed over three times the natural log of x, then we're going to have minus the integral of v du, which is going to be x cubed over three times one over x dx. So now that we've done that let's keep working through this. We have this x cubed over three natural log of x, then what I'm going to do is a couple steps at once here. I'm gonna cancel this x with one of these x's giving us an x squared on top, then I'll pull this one third out to the front. So I have minus one third that's going to be the integral, and then in this case, we'll have x squared d x. And this is definitely an integral we know how to solve using the power rule. So what this is gonna come out to be is x cubed over three l n x. That's going to be minus one third. That's going to be times the integral here, which is x cubed over three, then plus the constant of integration c. Then I can go ahead and distribute that one third into the brackets here giving us x cubed over three ln x minus, and then multiplying these together we'll have three times three, which is nine in the denominator, so x cubed over nine. Then we're going to have plus one third times c, which is all just gonna come out to another constant anyway. So I'll just keep that as the constant c, and this right here would be our solution. So that is how you can use integration by parts to solve this problem up here. Now remember, typically, when you see the logarithm, that's what you're gonna want to set equal to u. And then usually, it's going to be algebraic. Well, after that would be inverse and then algebraic. But keep in mind, this is the one case where one of the cases where you would not actually want to set your x squared term equal to the u immediately. Again, you could have tried that. And if you went through the process, you would have found that it didn't end up working or would lead you to an integral that didn't make sense. And that's always a good sign that you didn't set the right things equal to u and d b at the start. But since I knew that this x squared could be easily integrated and that the u would become much more simple with the natural log when we took its derivative, that allowed me to figure out what I needed to put in for these variables and allowed me to ultimately get to my solution. So hope you found this video helpful. Let me know if you have any questions, and let's move on.

Find the integral. ∫ x 2 ln ⁡ x d x \int_{}^{}x^2\ln xdx

x 3 3 ln ⁡ x − x 3 9 + C \frac{x^3}{3}\ln x-\frac{x^3}{9}+C

x 3 3 ln ⁡ x + x 3 9 + C \frac{x^3}{3}\ln x+\frac{x^3}{9}+C

x 3 3 ln ⁡ x + 1 x + C \frac{x^3}{3}\ln x+\frac{1}{x}+C

x 3 3 ln ⁡ x − 1 x + C \frac{x^3}{3}\ln x-\frac{1}{x}+C

12. Techniques of Integration

Integration by Parts

Calculus

12. Techniques of Integration

Integration by Parts

Struggling with Calculus?

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Multiple Choice

Find the integral.
$\int_{}^{} x^{2} \ln x d x$

$\frac{x^{3}}{3} \ln x + \frac{x^{3}}{9} + C$

$\frac{x^{3}}{3} \ln x + \frac{1}{x} + C$

$\frac{x^{3}}{3} \ln x - \frac{1}{x} + C$

$\frac{x^{3}}{3} \ln x - \frac{x^{3}}{9} + C$

Verified step by step guidance

Step 1: Recognize that the integral ∫x^2ln(x)dx requires the use of integration by parts. Recall the formula for integration by parts: ∫u dv = uv - ∫v du. Here, we need to choose u and dv appropriately.

Step 2: Let u = ln(x) and dv = x^2 dx. This choice is made because the derivative of ln(x) simplifies, and x^2 dx is easily integrable. Compute du = (1/x) dx and v = ∫x^2 dx = (x^3)/3.

Step 3: Substitute into the integration by parts formula: ∫x^2ln(x)dx = uv - ∫v du. Substituting u = ln(x), v = (x^3)/3, and du = (1/x) dx, we get: ∫x^2ln(x)dx = [(x^3)/3]ln(x) - ∫[(x^3)/3](1/x)dx.

Step 4: Simplify the remaining integral: ∫[(x^3)/3](1/x)dx = (1/3)∫x^2 dx. Compute this integral: ∫x^2 dx = (x^3)/3. Substituting back, we have: ∫x^2ln(x)dx = [(x^3)/3]ln(x) - (1/3)(x^3)/3.

Step 5: Combine terms and simplify the expression: [(x^3)/3]ln(x) - (x^3)/9. Finally, add the constant of integration C to account for the indefinite integral. The final expression is: (x^3)/3 ln(x) - (x^3)/9 + C.

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