February 24th, 2018, 01:03 PM  #1 
Member Joined: Apr 2017 From: New York Posts: 63 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: 6,454 Thanks: 567 
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: 7,236 Thanks: 2412 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: 63 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: 63 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: 63 Thanks: 6 
on line four*

February 24th, 2018, 02:44 PM  #7 
Member Joined: Apr 2017 From: New York Posts: 63 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: 63 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: 7,767 Thanks: 1017 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: 63 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 

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