February 24th, 2018, 01:03 PM  #1 
Member Joined: Apr 2017 From: New York Posts: 65 Thanks: 6  integration by parts xlny
Hi guys, Where do I make mistake. I have rewritten three times still getting the same wrong answer 
February 24th, 2018, 01:34 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,527 Thanks: 588 
Your attachment is difficult to read. Type it in directly.

February 24th, 2018, 02:16 PM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,313 Thanks: 2447 Math Focus: Mainly analysis and algebra 
Line 6: $\frac12y$ should be $+\frac12y$ But that probably doesn't matter as the integral in line 4 is wrong. It should be $\int y^2 + 2y \,\mathrm dy$. Last edited by v8archie; February 24th, 2018 at 02:19 PM. 
February 24th, 2018, 02:34 PM  #4 
Member Joined: Apr 2017 From: New York Posts: 65 Thanks: 6 
ok thanks checking quickly again. sorry for small picture. I am attaching better one ( I guess ) 
February 24th, 2018, 02:42 PM  #5 
Member Joined: Apr 2017 From: New York Posts: 65 Thanks: 6 
v times du = 1/2(y1) is that wrong? on line for to set up integral. 
February 24th, 2018, 02:42 PM  #6 
Member Joined: Apr 2017 From: New York Posts: 65 Thanks: 6 
on line four*

February 24th, 2018, 02:44 PM  #7 
Member Joined: Apr 2017 From: New York Posts: 65 Thanks: 6 
ok never mind yes line four is wrong Let me work on that again and send 
February 24th, 2018, 02:55 PM  #8 
Member Joined: Apr 2017 From: New York Posts: 65 Thanks: 6 
How is it now, please?

February 24th, 2018, 03:50 PM  #9 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,814 Thanks: 1046 Math Focus: Elementary mathematics and beyond 
$$y=1+x$$ $$dy\,=\,dx$$ $$\int(y1)\ln y\,dy=\int y\ln y\,dy\int\ln y\,dy$$ $$\int\ln y\,dy=y\ln y\int\,dy=y\ln yy+C$$ $$\int y\ln y\,dy=y(y\ln yy)\int y\ln yy\,dy=y(y\ln yy)\int y\ln y\,dy+\int y\,dy$$ $$2\int y\ln y\,dy=y(y\ln yy)+\frac{y^2}{2}+C$$ I'll let you finish up... 
February 24th, 2018, 04:42 PM  #10 
Member Joined: Apr 2017 From: New York Posts: 65 Thanks: 6 
Up to here I understood perfectly and liked your method very much indeed. could you explain what is u, and dv at this line: âˆ«ylnydy=y(ylnyâˆ’y)âˆ’âˆ«ylnyâˆ’ydy=y(ylnyâˆ’y)â ˆ’âˆ«ylnydy+âˆ«ydy I did not understand this part and I know in the last line I will substitute y=1+x but I didn't want to do by heart just wanted to learn 

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 04:18 PM 
Integration by parts  aaronmooy47  Calculus  6  April 2nd, 2017 01:33 PM 
Integration by Parts Help  fredlo2010  Calculus  8  September 2nd, 2014 12:49 PM 
Integration by parts  Zynoakib  Calculus  2  March 14th, 2014 05:21 PM 
Integration by parts  tnutty  Calculus  1  February 23rd, 2009 01:31 PM 