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Found in: Page 344

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# 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.

See the step by step solution

## 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, $$\int {{x^2}} \ln xdx = \frac{{{x^3}}}{3}\ln x - \frac{{{x^3}}}{9} + c$$