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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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    [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

1 Answers 1

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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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    @Kannappan Sampath: Thanks!2012-03-28