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

x33ln⁡x−x39+C\frac{$$x^3$$}{3}\ln x-\frac{$$x^3$$}{9}+C3x3lnx−9x3+C

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

in this problem, we are asked to find the integral of this function right here. Notice that we have theta times this cosine of two theta with respect to theta. Now something that I notice is we have the product of two functions, and there's not really a simple way to do a variable substitution or any rules that we've in the past been familiar with. So this is going to be a case right here where we're going to want to use integration by parts. Now the formula that we've used for integration by parts looks like this. The integral of u d v is equal to u v minus the integral of v d u. Now when using this formula, our first step is typically to figure out what u is going to be. Now recall that u is something that needs to become more simple when you take its derivative. Now if I look at what I have for the integral right here, I can see that we have just the theta right there. I know that if I take the derivative of theta, that's just gonna be one if it's with respect to theta. So So I'm going to say u equal to theta. That means that the derivative of u is going to be one d theta to show this is with respect to theta. So we've now found our u and our du. Now next, I need to find my dv. Now, dv needs to be something that can be easily integrated. Now, what about this cosine two theta term? Is this something that I can easily integrate? Well, I know how to integrate trig functions, so we should be able to integrate the cosine of two theta. But what exactly is that going to be? Well, let's figure this out. So if I integrate dv that's going to give me v, and to do that I have to integrate this cosine of two theta d theta. Now let's go ahead and work out what this integral is going to look like. So we'll have the integral of cosine two theta d theta, and what I first need to do is a variable substitution here. Notice here that we have two theta right here, I'm going to go ahead and set that equal to u. So what I'm going to say is that my u is two theta, and what that means is that du is going to be the derivative of two theta which would just be two with respect to theta. So now I've found u and du and what I can do is rewrite out this integral, so the cosine two theta d theta would be the same thing as the cosine of u replacing that two theta with a u, and then we have d theta. Now, that's going to be with my integral right there because we haven't done this integral step yet. Now, notice here that I'm trying to integrate a u with respect to theta, that's not really going to work, but notice that we said that du was equal to two d theta. So what I could do is write this as two d theta and then divide by that two right there to go ahead and get this to be the same expression that we have inside the integrand. So what I can do is move this two in the denominator outside to the front here, giving me one half the integral of cosine u, and that's going to be two d theta, which I can see right here is just du. So now we can integrate the cosine of u with respect to u, that's just gonna give us sine. So we're going to get one half and then the integral of cosine, well I know that that's just sine of u. And then what I can go ahead and do is take my two theta and plug it in right there to give me one half sine of two theta. So then and then we would have plus c when doing this integral right here, but I don't really need to write that plus c when this is just the portion that we're solving for integration by parts. So what I'm going to put in for this v here is one half sine two theta, and that's going to be the integral right there. So as you can see we have now gone ahead and found what our u and du are and what our dv and v are. So the last step would just be, or the the way to get this solution here would be to just plug everything into this portion of the equation. So what we're going to have is u times v, which is going to be theta times one half sine of two theta, and we're going to have minus the integral of v du, so we're going to have the minus, the integral right here of one half sine two theta with respect to theta. Now from this point, I can go ahead and simplify things, so I can write this all as theta over two sine two theta, and then from here we're going to have minus, and I'll pull that one half to the outside, then the integral of sine two theta. Now we know how to integrate these trig functions that we've seen before. We just need to go through the same process where we do a variable substitution, and then we go ahead and figure out our answer. Now I saw when I integrated this cosine of two theta, that gave me one half sine two theta. Now I know that the integral of sine turns out to be negative cosine. And so what I can do is recognize if I go through the variable substitution where I do a substitution right there, where where they make that u or w, that's just going to give me another one half on the outside and then whatever the integral of sine is, which I know is negative cosine. So rather than going through this whole process of doing the variable substitution strategically, I know that this whole thing is just gonna come out to minus one half. And the integral of sine two theta, well, I know that that's just going to be negative, and then we'll pull out this one half here, cosine of two theta, that's going to be plus the cost of integration c. Now at this point what I can do is distribute this one half into the brackets right about here. So doing that, we're gonna have this theta over two sine two theta, and we're going to have minus negative one half, which is gonna give me plus one fourth. So I have one fourth cosine two theta right about there, and then we'll have minus one half times c which would be negative one half c, but since c is just a constant, I'm just gonna continue to write this as plus c since that could be any constant that we have right there. So this right here is going to be my answer and the solution to this problem, and I need to write this parenthesis right here, so sine two theta. But that's gonna be the solution right there. So this is how you can do integration by parts when it comes to solving these more complicated types of problems. Hope you found this video helpful.

Find the integral.∫x2ln⁡xdx\int_{}^{}$$x^2$$\ln xdx∫x2lnxdx

x33ln⁡x−x39+C\frac{$$x^3$$}{3}\ln x-\frac{$$x^3$$}{9}+C3x3lnx−9x3+C

x33ln⁡x+x39+C\frac{$$x^3$$}{3}\ln x+\frac{$$x^3$$}{9}+C3x3lnx+9x3+C

x33ln⁡x+1x+C\frac{$$x^3$$}{3}\ln x+\frac{1}{x}+C3x3lnx+x1+C

x33ln⁡x−1x+C\frac{$$x^3$$}{3}\ln x-\frac{1}{x}+C3x3lnx−x1+C

11. Techniques of Integration

Integration by Parts

Business Calculus

11. Techniques of Integration

Integration by Parts

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객관식

Find the integral.
$$\int$_{}^{}x^2$\ln$ xdx$

$$\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$

검증된 단계별 안내

Step 1: Recognize that the integral involves a product of functions, x^2 and ln(x). This suggests using integration by parts, which is based on the formula: ∫u dv = uv - ∫v du.

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

Step 3: Compute du and v. Differentiate u to get du = (1/x) dx. Integrate dv to get v = (x^3)/3.

Step 4: Substitute into the integration by parts formula: ∫x^2 ln(x) dx = uv - ∫v du. Substituting the values, this becomes: ((x^3)/3) ln(x) - ∫((x^3)/3)(1/x) dx.

Step 5: Simplify the remaining integral. The term ∫((x^3)/3)(1/x) dx simplifies to (1/3) ∫x^2 dx. Integrate x^2 to get (x^3)/3. Combine all terms and add the constant of integration, C, to complete the solution.

8:23m

다음 시청

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Find the integral.∫x2lnxdx\(\int\)_{}^{}x^2\(\ln\) xdx∫x2lnxdx

다음 시청

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Find the integral.
$\(\int\)_{}^{}x^2\(\ln\) xdx$