Book edition 2nd

Author(s) James Stewart

Pages 830 pages

ISBN 9781133112280

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Short Answer

Evaluate the integral using integration by parts with the individual choices of u and dv
\(\;\;\;\;\)\(\;\)\(\)\(I = \int {{x^2}\ln xdx} \)
u=lnx dv=x²dx

Integration by parts by parts is performed as

\(\int {udv = uv - \int {vdu} } \)

We will take u=lnx and then find du/dx on substituting the values in the above formula we will get the required answer.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given Data

Given: I= \(\int {{x^2}\ln xdx} \)

U=lnx, dv=x²dx

Step 2: Evaluating the equation

u=Inx

du=dx/x

\(\int {dv = \int {{x^2}} } dx\)

V=x³/3

Step 3: Integrating the equation

\(\int {{x^2}} \ln xdx = \ln x.\frac{{{x^3}}}{3} - \int {\frac{{{x^3}}}{3}.\frac{{dx}}{x}} \)

= \(\int {\frac{{{x^3}}}{3}} \ln x - \int {\frac{{{x^3}}}{3}} \)

=\(\int {\frac{{{x^3}}}{3}} \ln x - \int {\frac{{{x^3}}}{9}} + c\)

Hence, \(\int {{x^2}} \ln xdx = \frac{{{x^3}}}{3}\ln x - \frac{{{x^3}}}{9} + c\)

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Essential Calculus: Early Transcendentals

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Short Answer

Evaluate the integral using integration by parts with the individual choices of u and dv
\(\;\;\;\;\)\(\;\)\(\)\(I = \int {{x^2}\ln xdx} \)
u=lnx dv=x²dx

Step by Step Solution

Step 1: Given Data

Step 2: Evaluating the equation

Step 3: Integrating the equation

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

Evaluate the integral using integration by parts with the individual choices of u and dv\(\;\;\;\;\)\(\;\)\(\)\(I = \int {{x^2}\ln xdx} \) u=lnx dv=x2dx

Step by Step Solution

Step 1: Given Data

Step 2: Evaluating the equation

Step 3: Integrating the equation

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Evaluate the integral using integration by parts with the individual choices of u and dv
\(\;\;\;\;\)\(\;\)\(\)\(I = \int {{x^2}\ln xdx} \)
u=lnx dv=x²dx