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Consider following integral: $$13\int{\frac{1}{8x-4}dx}\tag{1}$$ By factorizing the denominator and then taking the factor outside the integral sign, it can be rewritten as $$\frac{13}{4}\int{\frac{1}{2x-1}dx}\tag{2}$$

Now $(1)$ and $(2)$ should be equivalent, yet they evaluate into different integrals namely $$13\,\int{\frac{1}{8x-4}dx}=\frac{13}{8}\ln{|8x-4|}+C\tag{1a}$$ $$\frac{13}{4}\int{\frac{1}{2x-1}dx}=\frac{13}{8}\ln{|2x-1|}+C \tag{2a}$$

Since $(1)\equiv(2)$, then $(1a)\text{ and }(2a)$ should be equivalent as well, which reduces to $$\ln{|8x-4|}=\ln{|2x-1|}$$ which clearly isn't true. What am I missing here?

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    You are missing the constants!2012-03-28
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    @Nana Not sure I see what you mean2012-03-28
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    Surely, the two $C$'s can't be the same2012-03-28
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    Moreover, differentiating the two answers gives the same results.2012-03-28
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    The constants are different though, not $C$ in both cases. So if you see the answer here by Dennis, you'll see the constants are different therefore you can't cancel the constants and compare like that.2012-03-28
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    [Wikipedia's article on the constant of integration](https://secure.wikimedia.org/wikipedia/en/wiki/Constant_of_integration#Necessity_of_the_constant) has a section discussing a related but different example where you can't just disregard the constant.2012-03-28

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$$\begin{align}\frac{13}{8}\ln{|8x-4|}+C_1&=\frac{13}{8}\ln{(4|2x-1|)}+C_1 \\\ &=\frac{13}{8}\ln{|2x-1|}+\frac{13}{8}\ln4+C_1\\&=\frac{13}{8}\ln{|2x-1|}+C_2\end{align}$$

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    I have added an align environment to beautify the post. I hope it looks better now. +1!2012-03-28
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    @Kannappan Sampath: Thanks!2012-03-28