February 24th, 2018, 05:45 PM  #11 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,884 Thanks: 1088 Math Focus: Elementary mathematics and beyond  
February 24th, 2018, 06:39 PM  #12 
Senior Member Joined: Apr 2017 From: New York Posts: 119 Thanks: 6 
if u=y and dv= lny then du=dy and v=ylnyy perfect. thanks a lot sir. I couldn't figure out at first how v was now very clear. thanks million. 
February 24th, 2018, 07:17 PM  #13 
Senior Member Joined: Apr 2017 From: New York Posts: 119 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 09:20 PM. 
February 24th, 2018, 09:18 PM  #14 
Global Moderator Joined: Dec 2006 Posts: 19,951 Thanks: 1842 
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 $\displaystyle \begin{align*}\!\int\! x\ln(1 + x)\,dx &= \frac12\left(x^2 1\right)\ln(1 + x)  \frac12\!\int\! \frac{x^2  1}{1 + x}\, dx \\ &= \frac12\left(x^2  1\right)\ln(1 + x)  \frac12\!\int (x  1)\, dx \\ &= \frac12(x^21)\ln(1 + x)  \frac14(x  1)^2 + \text{C} \\ \end{align*}$ 

Tags 
integration, parts, xlny 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Integration By Parts  Leonardox  Calculus  4  February 22nd, 2018 05:18 PM 
Integration by parts  aaronmooy47  Calculus  6  April 2nd, 2017 02:33 PM 
Integration by Parts Help  fredlo2010  Calculus  8  September 2nd, 2014 01:49 PM 
Integration by parts  Zynoakib  Calculus  2  March 14th, 2014 06:21 PM 
Integration by parts  tnutty  Calculus  1  February 23rd, 2009 02:31 PM 