February 24th, 2018, 04:45 PM  #11 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,767 Thanks: 1017 Math Focus: Elementary mathematics and beyond  
February 24th, 2018, 05:39 PM  #12 
Member Joined: Apr 2017 From: New York Posts: 63 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, 06:17 PM  #13 
Member Joined: Apr 2017 From: New York Posts: 63 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: 18,719 Thanks: 1536 
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 $\displaystyle \begin{align*}\!\int\! x\ln(1 + x)\,dx &= \frac12\left(x^2 1\right)\ln(1 + x)  \frac12\!\int\! \frac{x^2  1}{1 + x}\, dx \\ &= \frac12\left(x^2  1\right)\ln(1 + x)  \frac12\!\int (x  1)\, dx \\ &= \frac12(x^21)\ln(1 + x)  \frac14(x  1)^2 + \text{C} \\ \end{align*}$ 

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integration, parts, xlny 
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