Integration - involving logarithm

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I have a slight confusion on how to integrate functions of the form: $$\int\frac{a}{x}dx$$

Suppose we have the following function: $$\int\frac{-2}{x}dx$$ There are two ways we can proceed to integrate this function. One is to treat the $-2$ sign as a constant and take it out of the integration function: $$-2\int\frac{1}{x} = -2\ln{x} = \ln{\frac{1}{x^2}}$$ Another way to do this is to treat the $-2$ as part of the integration variable: $$\int\frac{1}{-0.5x}dx = -2\ln{(-0.5x)}$$

This seems to obtain two different answers. Which method is correct? Or is there something that I'm missing?

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To be proper, you should use that

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

not just $\ln x$ (as defining the logarithm for negative numbers requires some care). Given this, you found two antiderivatives:

$$\ln \frac{1}{x^2} \quad \text{ and } -2 \ln (.5 |x|)$$

The second one can be written as

$$-2 \ln (.5) - 2 \ln |x| = -2 \ln (.5) + \ln \frac{1}{x^2}$$

These differ by a constant, which is not a problem.

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Both methods are correct. The results differ by a constant:

$$-2\ln|{-0.5\cdot x}|=-2\ln|x|-2\ln(0.5)=-2\ln|x|+C$$

Keep in mind that $$\int \frac1x \,dx = \ln|x|+C$$