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Integrate the functions.$$x(\log x)^{2}$$

Calculus 2 / BC

Chapter 7

Integrals

Section 6

Integration by Parts

Integration Techniques

Harvey Mudd College

University of Nottingham

Boston College

Lectures

01:53

In mathematics, integratio…

27:53

In mathematics, a techniqu…

02:05

Evaluate the integral.…

02:01

Find the integrals.$$\…

03:44

evaluate using Integration…

04:02

Evaluate the integrals.

0:00

Evaluate the integral. $\i…

06:19

02:24

00:47

Evaluate the indefinite in…

00:34

00:33

03:14

Find the antiderivative.

Find the antiderivative. <…

03:20

02:27

03:47

05:13

06:04

04:48

05:52

Evaluate $\int \frac{(\ln …

03:31

We have problem # 14 in which we need to integrate X log x whole square. So we'll be using integration by parts. Since this is logarithmic function let's take as first function. This is second function. So according to formula for integration by parts. First function log x whole square integration of second function minus differentiation of first function, integration of second function and dx. Okay, so This is long. Exhale Square in to access acquired by 2-. Uh This differentiation will be too Log ax into differentiation of low taxes. One by X into its integration is access acquired by two and DX two and 2 will get cancelled out X&X will get cancelled out. So if you're left with Access required by two, log X whole square minus. Yeah log X into X. Dx. So one second we have to use integration by part four. This so we'll be using access acquired by two first letters right minus first function, second function so into uh first function letters right? First function integration of second function minus differentiation of first function. Integration of second function and call integration. So this is access acquired by two log x whole square minus This will be x. so x esquire by two Log ax. And plus this is access choir by two. And differentiation of log oxygen by X. Access required by two DX XX will get cancelled out. So finally access acquired by two love X whole square minus X squared by two log X plus one by two. Integration of X will be Access required by two plus the Okay, so we will be having The integration has access acquired by two log ax hole square minus X squared by two log x. Blessed one by four X squared plus C. They should be the answer. Thank you so much.

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