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Essential Calculus: Early Transcendentals
Found in: Page 344
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

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

  1. \(\;\;\;\;\)\(\;\)\(\)\(I = \int {{x^2}\ln xdx} \)

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.

See the step by step solution

Step by Step Solution

Step 1: Given Data

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

U=lnx, dv=x2dx

Step 2: Evaluating the equation



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


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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