StudySmarter AI is coming soon!

- :00Days
- :00Hours
- :00Mins
- 00Seconds

A new era for learning is coming soonSign up for free

Suggested languages for you:

Americas

Europe

Q1E

Expert-verifiedFound in: Page 344

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=x ^{2}dx**

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.

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

U=lnx, dv=x^{2}dx

u=Inx

du=dx/x

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

V=x^{3}/3

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

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

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

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

94% of StudySmarter users get better grades.

Sign up for free