My Math Forum integration by parts xlny

 Calculus Calculus Math Forum

February 24th, 2018, 04:45 PM   #11
Global Moderator

Joined: Oct 2008
From: London, Ontario, Canada - The Forest City

Posts: 7,859
Thanks: 1080

Math Focus: Elementary mathematics and beyond
Quote:
 Originally Posted by greg1313 $$\int y\ln y\,dy=y(y\ln y-y)-\int y\ln y-y\,dy=y(y\ln y-y)-\int y\ln y\,dy+\int y\,dy$$
$$u=y$$

$$dv=\ln y$$

Ok?

 February 24th, 2018, 05:39 PM #12 Member   Joined: Apr 2017 From: New York Posts: 92 Thanks: 6 if u=y and dv= lny then du=dy and v=ylny-y perfect. thanks a lot sir. I couldn't figure out at first how v was now very clear. thanks million. Thanks from greg1313
 February 24th, 2018, 06:17 PM #13 Member   Joined: Apr 2017 From: New York Posts: 92 Thanks: 6 From that point I will use another version as u=lny and dv=y .... do you recommend that path or ? I know you used u=y and dv=lny, but I was not practical integrating lny and couldn't see the result fast. Last edited by skipjack; February 24th, 2018 at 08:20 PM.
 February 24th, 2018, 08:18 PM #14 Global Moderator   Joined: Dec 2006 Posts: 19,527 Thanks: 1750 Let $u = \ln(1 + x)$ and $v = \frac12\left(x^2 - 1\right)$, so that $\displaystyle du = {\small\frac{1}{1 + x}}\,dx$ and $dv = x\, dx$, then \displaystyle \begin{align*}\!\int\! x\ln(1 + x)\,dx &= \frac12\left(x^2- 1\right)\ln(1 + x) - \frac12\!\int\! \frac{x^2 - 1}{1 + x}\, dx \\ &= \frac12\left(x^2 - 1\right)\ln(1 + x) - \frac12\!\int (x - 1)\, dx \\ &= \frac12(x^2-1)\ln(1 + x) - \frac14(x - 1)^2 + \text{C} \\ \end{align*} Thanks from greg1313

 Tags integration, parts, xlny

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Leonardox Calculus 4 February 22nd, 2018 04:18 PM aaronmooy47 Calculus 6 April 2nd, 2017 01:33 PM fredlo2010 Calculus 8 September 2nd, 2014 12:49 PM Zynoakib Calculus 2 March 14th, 2014 05:21 PM tnutty Calculus 1 February 23rd, 2009 01:31 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top