Proof In Exercises 71-76, use integration by parts to prove the formula. (For Exercises 71-74 ,

Proof In Exercises 71-76, use integration by parts to prove...

Key Concepts

Integration by Parts

Integration by parts is a method of integration based on the product rule for differentiation. It transforms the integral of a product of functions into a potentially easier form by choosing one function to differentiate and another to integrate, summarized by the formula ? u dv = uv - ? v du.

Selection of u and dv

Choosing which part of the integrand to designate as u and which as dv is crucial for simplifying the integration process. Often, functions that become simpler when differentiated (like logarithmic functions) are chosen for u, while functions that are easy to integrate (like polynomials) are selected for dv.

Handling of Indefinite Integrals

In indefinite integration, the result includes an arbitrary constant of integration 'C' since antiderivatives differ by a constant. This reflects the fact that integration is the inverse process of differentiation, which loses constant terms.

Verification by Differentiation

Once an antiderivative is obtained, differentiating it to obtain the original integrand serves as an important check. This process relies on the use of differentiation rules such as the product and chain rules to ensure that the integration was performed correctly.

Transcript

00:01 Again, we're using integration by parts to show that this left -hand side is exactly the same as the right -hand side, right here.

00:09 So we're going to still use our leight because it's integration by parts.

00:17 And we're going to decide which of these two functions here at this left -hand side.

00:26 These two functions here should be our u and rdv.

00:30 Le8, this is the 8.

00:37 So this is logarithm, this is inverse trigonometric, and this is algebra.

00:43 You can see that it's a combination or is a product of logarithm and algebra.

00:49 So the logarithm takes precedence over the algebra in terms of deciding which of the functions to take the u.

00:57 So because of that, then the u is going to be the logarithm and then the the dv is going to be the algebraic so this is going to be our u as perliate and this is going to be our dv so we let u be logarithm of x and then dv be x power n then what is going to be d u d u is 1 over x d x d x d x that's a derivative of this logarithm and then v you know the definition for integration is x so whatever power you have here you just add one more then you divide by the new power new power is that okay so whatever is here add one more and then you divide by the new power which is n plus one that is v and then the integration by parse formula, i'm going to invoke it.

02:13 So it's just udv, which is uv, uv minus integral v, du.

02:26 So you have, that is uv.

02:31 So what is u, u is natural algorithm of x? v, what is v? v is x to the n plus 1 over n plus 1.

02:44 And then minus integral v, du.

02:50 So what is v? v is x to the n plus 1 over n plus 1, and du is 1 over x dx, so multiply by 1 over x dx.

03:06 Great.

03:08 So we have to evaluate this integral again.

03:13 So let's actually do it somewhere here.

03:21 Let's do this one over here.

03:25 So, we're doing it here.

03:29 So what is this integral? x n plus 1 over n plus 1 over x d x.

03:37 Now, this x n plus 1 here, okay, is the same as this x n plus 1 here, like x to the end.

03:54 I'm trying to do everything on the same page.

03:57 Sometimes i do not like to go to a different page because i have to toggle back and forth between pages.

04:07 But this x to the n plus 1 here is the same as indices, by lots of indices, the same as x to the n times x to the 1.

04:21 Okay, so i'm going to change this one to x to n times x to the 1.

04:27 So this is the same as x to the n times x.

04:37 X to the 1 is still x.

04:39 Then this x is going to cancel that x.

04:43 And this is just a constant because the integral is with respect to x, and this is just n plus 1.

04:50 There's no x here.

04:52 So it's a constant, and i'm going to pull it out, and then this is just going to be x to the ntx...

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