Evaluate $\int\frac{\mathrm{d}x}{2x-4}$

Question

My question is to evaluate:

$$\int\frac{\mathrm{d}x}{2x-4}$$

Why is the solution equal to $\frac{1}{2}\ln|x-2|$ as opposed to $\frac{1}{2}\ln|2x-4|$?

I understand that if I factor $\frac{1}{2}$ before integrating I would get the first answer, but if I don't factor before and just use $u$ substitution, why would the second answer be wrong?

Answer 1

'Similar' approach as yours bangsauce. Let $u=2x-4\;\Rightarrow\; du=2\ dx\;\Rightarrow\;du=\dfrac12\ dx$, then $$ \begin{align} \int\frac{dx}{2x-4}&=\frac12\int\frac{du}{u}\\ &=\frac12\ln|u|+C\\ &=\frac12\ln|2x-4|+C\quad\Rightarrow\quad\text{your solution}\\ &=\frac12\ln2|x-2|+C\\ &=\frac12\ln|x-2|+\frac12\ln 2+C\quad\Rightarrow\quad\ln ab=\ln a +\ln b\\ &=\frac12\ln|x-2|+K\quad\Rightarrow\quad\text{where }K=\frac12\ln 2+C\text{ , yield another solution}. \end{align} $$

Answer 2

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

i.e., $$\frac12\ln|2x-4|-\ln|x-2|=\ln2 $$ which is a constant

So, $$\frac{d\left(\frac12\ln|2x-4|\right)}{dx}=\frac{d\left(\frac12\ln|x-2\right)}{dx}$$

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Alternatively, $$\frac12\ln|2x-4|+k=\frac12\left(\ln 2+\ln|x-2|\right)=k'+\frac12\ln|x-2|$$ where $$k'=k+\frac12\ln2$$