Logarithmic inconsistency when integratingConsider following integral:13∫18x−4dx&nbsp;(1)&nbsp;By factorizing the denominator and then taking the factor outside the integral sign, it can be rewritten as134∫12x−1&nbsp;dx&nbsp;&nbsp;(2)Now (1) and (2) should be equivalent, yet they evaluate into different integrals namely13∫18x−4&nbsp;dx&nbsp;=138ln|8x−4|+C&nbsp;(1a)134∫12x−1&nbsp;dx&nbsp;=138ln|2x−1|+C&nbsp;(2a)Since&nbsp;(1)≡(2)&nbsp;, then (1a) and (2a) should be equivalent as well, which reduces toln|8x−4|=ln|2x−1|which clearly isn't true. What am I missing here?

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Logarithmic inconsistency when integratingConsider following integral:13∫18x−4dx&amp;nbsp;(1)&amp;nbsp;By factorizing the denominator and then taking the factor outside the integral sign, it can be rewritten as134∫12x−1&amp;nbsp;dx&amp;nbsp;&amp;nbsp;(2)Now (1) and (2) should be equivalent, yet they evaluate into different integrals namely13∫18x−4&amp;nbsp;dx&amp;nbsp;=138ln|8x−4|+C&amp;nbsp;(1a)134∫12x−1&amp;nbsp;dx&amp;nbsp;=138ln|2x−1|+C&amp;nbsp;(2a)Since&amp;nbsp;(1)≡(2)&amp;nbsp;, then (1a) and (2a) should be equivalent as well, which reduces toln|8x−4|=ln|2x−1|which clearly isn&#039;t true. What am I missing here?

Wilson Rivas · Accepted Answer

138ln⁡{|8x−4|}+C1=138ln⁡{(4|2x−1|)}+C1=138ln⁡{|2x−1|}+138ln⁡4+C1=138ln⁡{|2x−1|}+C2

Logarithmic inconsistency when integrating Consider following integral: \(\displaystyle{13}\int{\frac{{{1}}}{{{8}{x}-{4}}}}{\left.{d}{x}\right.}{t}{a}{g}{1}\) (1) By

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