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Using u-substitution I'm able to integrate the following two integrands:

$$\int \frac{1}{2x+3}dx = \frac{1}{2}\ln|2x+3|+C$$

$$\int \frac{1}{2x+5}dx = \frac{1}{2}\ln|2x+5|+C$$

Following this pattern I would assume:

$$\int \frac{1}{2x+4}dx = \frac{1}{2}\ln|2x+4|+C$$

However when I enter this expression into wolfram alpha I get:

$$\int \frac{1}{2x+4}dx = \frac{1}{2}\ln|x+2|+C$$

I can even arrive at this result myself by factoring out the $\frac{1}{2}$ before integration like this:

$$\int \frac{1}{2}\cdot\frac{1}{x+2}dx = \frac{1}{2}\int \frac{1}{x+2}dx$$

But I don't understand how this can be the case? How can two different techniques give me different results? Is there a property of logs that I don't understand?

I would think that

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

Or are they actually equal somehow? PLEASE help me understand what's going on here.

NotADeveloper
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Nick
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1 Answers1

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Because $$\frac{1}{2}\ln|2x+4|+C=\frac{1}{2}\ln|x+2|+\frac{1}{2}\ln2+C,$$ where $\frac{1}{2}\ln2+C$ is a constant.

Michael Rozenberg
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