My Math Forum Integration By Parts

 Calculus Calculus Math Forum

February 22nd, 2018, 03:50 PM   #1
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Joined: Apr 2017
From: New York

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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.
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 February 22nd, 2018, 03:54 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,963 Thanks: 1148 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: 3,017 Thanks: 1603 $x \color{red}{\ln(3x+1)} -x+\dfrac{1}{3} \color{red}{\ln(3x+1)} + C$ factor out $\color{red}{\ln(3x+1)}$ from the two terms ... $\color{red}{\ln(3x+1)}\left[x + \dfrac{1}{3}\right]- x + C$ $\color{red}{\ln(3x+1)}\left[\dfrac{3x}{3} + \dfrac{1}{3}\right]- x + C$ $\color{red}{\ln(3x+1)}\left[\dfrac{3x+1}{3}\right]- x + C$ $\dfrac{1}{3}\ln(3x+1) \cdot (3x+1) - x + C$ Thanks from topsquark
 February 22nd, 2018, 04:17 PM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,963 Thanks: 1148 Math Focus: Elementary mathematics and beyond $\displaystyle x\ln(3x+1)-x+\frac13\ln(3x+1)+C$ $\displaystyle \frac13(3x\ln(3x+1)+\ln(3x+1))-x+C$ $\displaystyle \frac13(3x+1)\ln(3x+1)-x+C$ Thanks from topsquark
 February 22nd, 2018, 04:18 PM #5 Senior Member   Joined: Apr 2017 From: New York Posts: 155 Thanks: 6 whooops yobbaaa, I thought and tried everything but factoring out. thanks guys really appreciate. Thanks from greg1313

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