Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 7 - Section 7.1 - Integration by Parts - 7.1 Exercises - Page 477: 42


$\ln x\cdot \arcsin(\ln x) +\sqrt{1-(\ln x)^{2}} +C$

Work Step by Step

Substitute $\left[\begin{array}{ll} t=\ln x, & x=e^{t}\\ dt=\frac{dx}{x} & \end{array}\right]$ $I=\displaystyle \int\frac{\arcsin(\ln x)}{x}dx=\int\arcsin tdt$ By parts, we take $\displaystyle \left[\begin{array}{ll} u=\arcsin t & dv=dt\\ & \\ du=\frac{1}{\sqrt{1-t^{2}}} & v=t \end{array}\right],\quad\int udv=uv-\int vdu$ $I=t\displaystyle \arcsin t-\int\frac{t}{\sqrt{1-t^{2}}}dt$ $I_{1}=\displaystyle \int\frac{tdt}{\sqrt{1-t^{2}}}$= substitute $\left[\begin{array}{l} w=1-t^{2}\\ dw=-2tdt \end{array}\right]$ = $-\displaystyle \frac{1}{2}\int w^{-1/2}dt=-\frac{1}{2}(\frac{x^{1/2}}{1/2})$ $=-w^{1/2}+C$ $=-\sqrt{1-t^{2}}$ $I=t\arcsin t+\sqrt{1-t^{2}}+C$ ... bring back x... $I=\ln x\cdot \arcsin(\ln x) +\sqrt{1-(\ln x)^{2}} +C$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.