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Problem:

$$\int\frac{1}{2}-\frac{3}{2x+4}dx$$

Using two different methods I am getting two different answers and have trouble finding why.

Method 1:

$$\int\frac{1}{2}-\frac{3}{2x+4}dx$$

$$\int\frac{1}{2}-\frac{3}{2}\left(\frac{1}{x+2}\right)dx$$ $$\int\frac{1}{2}dx-\frac{3}{2}\int\frac{1}{x+2}dx$$ $$\frac{1}{2}x-\frac{3}{2}\int\frac{1}{x+2}dx$$ $$x+2=u$$ $$dx=du$$ $$\frac{1}{2}x-\frac{3}{2}\int\frac{1}{u}du$$

$$\frac{1}{2}x-\frac{3}{2}\ln{|x+2|}+C$$

Method 2: $$\int\frac{1}{2}-\frac{3}{2x+4}dx$$

$$\int\frac{1}{2}-3\left(\frac{1}{2x+4}\right)dx$$ $$\int\frac{1}{2}dx-3\int\frac{1}{2x+4}dx$$ $$\frac{1}{2}x-3\int\frac{1}{2x+4}dx$$ $$2x+4=u$$ $$dx=\frac{du}{2}$$ $$\frac{1}{2}x-\frac{3}{2}\int\frac{1}{u}du$$

$$\frac{1}{2}x-\frac{3}{2}\ln{|2x+4|}+C$$

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    $\begingroup$ Note that $\left| 2x+4 \right| = 2 \left| x+2 \right|$ and then use the properties of logarithms to see that the factor of 2 is essentially just a constant in addition. $\endgroup$ – Matti P. Aug 13 at 11:44
  • $\begingroup$ As others have mentioned, your issue is resolved by a logarithm property. ... Often, the trickiest part of learning Calculus is remembering your Pre-Calculus. :) $\endgroup$ – Blue Aug 13 at 11:48
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    $\begingroup$ This type of question, like find the difference between $\int (x+1 )dx=\frac{x^2}{2}+x+C$ and $\int (x+1)dx=\frac{(x+1)^2}{2}+C$, appears here so frequently, and the answer is always the same (there are no difference, only the constants differs). Can a canonical answer be made for it? $\endgroup$ – Zacky Aug 13 at 11:52
  • $\begingroup$ Further to @Zacky's point, in some cases it's as simple as "they're clearly the same" (up to a constant or otherwise), but sometimes proving the equivalence requires e.g. a trigonometric identity the OP doesn't know how to prove. I'm not sure whether that means we should have multiple "canonical" versions or just give up on trying it altogether. $\endgroup$ – J.G. Aug 13 at 12:59
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Both answers are the same, mind the constant: $$\ln|2x+4|+\color{blue}{C_1}=\ln|2\left(x+2\right)|+\color{blue}{C_1}=\ln|x+2|+ \underbrace{\ln 2 + \color{blue}{C_1}}_{\color{purple}{C_2}} = \ln|x+2|+\color{purple}{C_2}$$

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$$\begin{align}\frac12 x + \frac 32 \ln |2x + 4| + C_1 &= \frac12x + \frac32\ln(2\cdot|x+2|) + C_1\\&=\frac12x + \frac32(\ln 2 + \ln|x+2|)+C_1\\&=\frac12x + \frac32 \ln|x+2| + (\frac 32 \ln 2 + C_1)\\&=\frac12x + \frac32\ln|x+2| + C_2\end{align}$$

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$\ln|2x+4|=\ln|2(x+2)|=\ln|2|+\ln|x+2|=\ln|x+2|+C$

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More generally for positive real $a,b,d$

$$\ln{\big|da+db\big|}=\ln{\big|d(a+b)\big|}=\ln{\big|d\big|}+\ln{\big|a+b\big|}=c_1+\ln{\big|a+b\big|}$$

where $a=x,b=2,d=2$ in your example. Then, if $e=-\frac{3}{2}$ and $f=\frac{1}{2}$,

$$fa+\ln{\big|da+db\big|}+c_2=fa+\ln{\big|a+b\big|}+c_1+c_2=fa+\ln{\big|a+b\big|}+c_3$$

so the difference is only by a constant.

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