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February 22nd, 2018, 03:50 PM   #1
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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
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Factor a 1/3 out of I. That should get you started.
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February 22nd, 2018, 04:01 PM   #3
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$x \color{red}{\ln(3x+1)} -x+\dfrac{1}{3} \color{red}{\ln(3x+1)} + C$

factor out $\color{red}{\ln(3x+1)}$ from the two terms ...

$\color{red}{\ln(3x+1)}\left[x + \dfrac{1}{3}\right]- x + C$

$\color{red}{\ln(3x+1)}\left[\dfrac{3x}{3} + \dfrac{1}{3}\right]- x + C$

$\color{red}{\ln(3x+1)}\left[\dfrac{3x+1}{3}\right]- x + C$

$\dfrac{1}{3}\ln(3x+1) \cdot (3x+1) - x + C$
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February 22nd, 2018, 04:17 PM   #4
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$\displaystyle x\ln(3x+1)-x+\frac13\ln(3x+1)+C$

$\displaystyle \frac13(3x\ln(3x+1)+\ln(3x+1))-x+C$

$\displaystyle \frac13(3x+1)\ln(3x+1)-x+C$
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February 22nd, 2018, 04:18 PM   #5
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whooops yobbaaa,
I thought and tried everything but factoring out.

thanks guys really appreciate.
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