February 22nd, 2018, 03:50 PM  #1 
Member Joined: Apr 2017 From: New York Posts: 65 Thanks: 6  Integration By Parts
Hi guys. Can somebody explain me only the last line ( in red frame). how did the last line shape just after the previous one. I know the topic well just didn't get that algebra part of ln transformation. appreciate it. 
February 22nd, 2018, 03:54 PM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,814 Thanks: 1046 Math Focus: Elementary mathematics and beyond 
Factor a 1/3 out of I. That should get you started.

February 22nd, 2018, 04:01 PM  #3 
Math Team Joined: Jul 2011 From: Texas Posts: 2,755 Thanks: 1405 
$x \color{red}{\ln(3x+1)} x+\dfrac{1}{3} \color{red}{\ln(3x+1)} + C$ factor out $\color{red}{\ln(3x+1)}$ from the two terms ... $\color{red}{\ln(3x+1)}\left[x + \dfrac{1}{3}\right] x + C$ $\color{red}{\ln(3x+1)}\left[\dfrac{3x}{3} + \dfrac{1}{3}\right] x + C$ $\color{red}{\ln(3x+1)}\left[\dfrac{3x+1}{3}\right] x + C$ $\dfrac{1}{3}\ln(3x+1) \cdot (3x+1)  x + C$ 
February 22nd, 2018, 04:17 PM  #4 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,814 Thanks: 1046 Math Focus: Elementary mathematics and beyond 
$\displaystyle x\ln(3x+1)x+\frac13\ln(3x+1)+C$ $\displaystyle \frac13(3x\ln(3x+1)+\ln(3x+1))x+C$ $\displaystyle \frac13(3x+1)\ln(3x+1)x+C$ 
February 22nd, 2018, 04:18 PM  #5 
Member Joined: Apr 2017 From: New York Posts: 65 Thanks: 6 
whooops yobbaaa, I thought and tried everything but factoring out. thanks guys really appreciate. 

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