Evaluate the indefinite integral. ∫lnxx4dx\int^{}\frac{\ln x}{x^4}dx

− ln ⁡ x 3 x 3 − 1 9 x 3 + C -\frac{\ln x}{3x^3}-\frac{1}{9x^3}+C

Evaluate the indefinite integral. ∫ ln x x 4 d x \int^{}\frac{\ln x}{x^4}dx

in this example, we're asked to evaluate the following integral. Now notice we have the integral of the natural log of x over x to the fourth power. Now, I'm going to write this in a little bit of a different way to hopefully allow us to see this integral a little more clearly. So this would be the same thing as saying we have one over x to the fourth power times the natural log of x. Now this hopefully shows us in a more clear way that we have the product of two functions that we're dealing with in this integral, And as I can see, I have a natural log right here in my integral and really no way of doing a sort of variable substitution to make this integral any more simple. So this is a case where we will want to use integration by parts. Now the way I'm going to apply integration by parts here is by creating a table. So in this table we're going to have two columns here, one for the derivatives and one for the integrals. And keep in mind the column for the derivatives is just what we would set equal to u, and the column for the integrals is what would be our dv. Now what exactly should be my u? Well, u should be more simple as we take the derivative. Now you may recall that it's usually best to actually set the natural log to the derivative, because if I went ahead and or set that equal to u I should say, because if I take derivatives of the natural log, I know that's going to become one over x, which is more simple. However, if I were to set this equal to to u, if I were to take these derivatives it wouldn't necessarily become more simple. So because of this, I'm going to say that u is going to be the natural log of x. Now keep in mind, something in general when you're setting the u and d v is to take a look and see what's gonna become more simple and what can be easily integrated as well. I know that the natural log is not something that can be easily integrated. So in this case we're saying that u is our natural log of x, and then if I take the derivative of the natural log of x that's going to give me one over x. I'm going to go ahead and stop right here at one over x, I'm going to stop taking derivatives for now, and let's now figure out what our dv should be. Well, dv is going to be whatever we can easily integrate. Let's see we have this one over x to the fourth power. I know that that is something I can use the power rule on when integrating. So I'm going to say one over x to the fourth power is what I'm gonna put in the integration column, or dv. So now we have the derivative right here of the natural log one over x, and what I'm going to now do is integrate one over x to the fourth power. How do I do that? Well integrating one over x to the fourth power is the same thing as integrating x to the negative fourth power. What I can do is use the power rule, So that's where I add one to this exponent and then divide by that exponent. Doing this is going to give me, and I can go ahead and clear this since we're saying this equal to this, so doing this integral over here is going to give me x to the negative three over negative three plus c, which I can go ahead and rewrite as negative one over three x times x cubed plus constant integration c. So in order to integrate this, what that's going to give me is negative one over three x cubed. So now we have this. And one more thing that I wanna go ahead and put on here as I'm doing this this table is you wanna make sure you're alternating the signs on your column for the derivative. So this one will be positive, and that one will be negative. Then you go positive, negative, positive, negative. We can go ahead and stop right there. Okay. So now we have this all written out. So what can I do from here? Well, what I now need to do is start with the first thing I have in my derivative column right here. I need to go diagonally down into the right and multiply there. So what I'm going to have is that this whole integral is equal to the natural log of x that's going to be multiplied by negative one over three x to the third power, and we're going to have in this case we'll move to this term so it's going to be minus, and we're going to multiply these directly across. Now notice here how since I'm multiplying directly across I'm no longer going diagonal I have to put that in an integral. So that's going to be one over x, that's going to be times negative one over three x cubed. And what I can do is also recognize this negative sign will cancel with that one, so we can just go ahead and add this integral on. So I'm gonna rewrite this. So we're gonna have negative l n of x putting this on top right here, and then gonna divide this by three x to the third power or three x cubed, I should say. Then we'll have plus and then this integral. And what I can do is multiply this x by the x in the denominator here. So that's gonna be and then I'll pull this actually, so that's gonna be one third, which I'll pull outside here, and then we're going to have this multiplied by that, which is going to give me one over x to the fourth power. This is just one over three x to the fourth power that we end up right there. Then we're integrating that with respect to x. So now I need to figure out what the integral of x of one over x to the fourth power is. Well, we already know what that is. That's just gonna be negative one over three x cubed. So if I go ahead and do this math here we're going to have negative natural log of x over three x cubed, and that's going to be plus, and we're going to have one third times, and in this case it's going to be the integral of one over x to the fourth power which is negative one over three x cubed, is going to be plus the constant of integration c. Now as the last step, we can go ahead and distribute this one third into these brackets here. So we're going to keep this natural log of x over three x cubed right about there, and we're going to have plus, and it's going to be one third times negative one third, and then x cubed in the denominator, which is going to be the same thing as minus one over nine x cubed. Then we're going to have plus the big constant c combining this constant with that one, and that's going to be the solution to our problem. That's how we can do integrated integration by parts using this table method. So hope you found this video helpful. This is the general strategy. As you can see, a lot of times, what we're really just trying to do is figure out how we're going to get this product to be something that's more simple or something that we can easily integrate, which is why we continue taking derivatives and integrals until we can go ahead and get this pattern down, something that we can integrate and then get our solution. So that's how you can solve these problems. Let me know if you have any questions.

Evaluate the indefinite integral. ∫ ln ⁡ x x 4 d x \int^{}\frac{\ln x}{x^4}dx

− ln ⁡ x 3 x 3 − 1 9 x 3 + C -\frac{\ln x}{3x^3}-\frac{1}{9x^3}+C

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

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

ln ⁡ x 3 x 3 − 1 9 x 3 + C \frac{\ln x}{3x^3}-\frac{1}{9x^3}+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

Evaluate the indefinite integral.
$\int^{} \frac{\ln x}{x^{4}} d x$

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

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

$\frac{\ln x}{3 x^{3}} - \frac{1}{9 x^{3}} + C$

$- \frac{\ln x}{3 x^{3}} - \frac{1}{9 x^{3}} + C$

Verified step by step guidance

Step 1: Recognize that the integral involves a combination of a logarithmic function (ln x) and a power of x. The integral to evaluate is ∫(ln x)/(x^4) dx.

Step 2: Use integration by parts. Recall the formula for integration by parts: ∫u dv = uv - ∫v du. Here, let u = ln x (since its derivative simplifies) and dv = x^(-4) dx (since its integral is straightforward).

Step 3: Compute du and v. Differentiate u = ln x to get du = (1/x) dx. Integrate dv = x^(-4) dx to get v = -1/(3x^3).

Step 4: Substitute into the integration by parts formula. Using ∫u dv = uv - ∫v du, substitute u = ln x, v = -1/(3x^3), and du = (1/x) dx. This gives: (-ln x)/(3x^3) - ∫[-1/(3x^3) * (1/x)] dx.

Step 5: Simplify the remaining integral. The term inside the integral becomes -1/(3x^4). Integrate this term to get (1/9x^3). Combine all terms to write the final result as: -ln x/(3x^3) - 1/(9x^3) + C.

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