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Reasoning for $\frac{\ln|x|}{2}+ C$ :

$$\int \frac{1}{(2x)} \, dx = \frac{\ln|2x|}{2}+ C$$

Check: By taking derivative we get $\frac{1}{x}$

Reasoning for $\frac{\ln|x|}{2}+ C$ :

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

Check: By taking derivative we get $\frac{1}{x}$

Question: Is $\int \frac{1}{2x} \, dx $ equal to $\frac{\ln|2x|}{2}+ C$ or $\frac{\ln|x|}{2}+ C$?

Sil
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  • They are equivalent by one line of algebra. –  Feb 27 '16 at 00:42
  • Choose whichever you prefer; they're the same. – egreg Feb 27 '16 at 00:45
  • You might understand that you can choose both if you remind yourself that $C$ is an indeterminate. – choco_addicted Feb 27 '16 at 00:57
  • The equals sigh in $\int f=g+C$ is not to be taken literally. It says that g is an anti-derivative of f, not THE anti-derivative of f. – DanielWainfleet Feb 27 '16 at 01:17
  • Find the derivative of both functions and compare your answers. It's important that you understand that you can always satisfy yourself that you have found the integral by differentiating the function you found. – MasB Feb 27 '16 at 02:49

2 Answers2

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Notice that $$ \frac{\ln|2x|}{2} = \frac{\ln|2|}{2}+\frac{\ln|x|}{2}$$ and since integration is unique up to constant, resulting integrals are the same. In other words your two results are differing only by constant, thus giving same indefinite integral.

Keep in mind that formally result of indefinite integral is set of functions rather than one specific function. You can think of the integral as a set $$\int f(x) dx = \{F(x); F'(x)=f(x)\}.$$ Then you can quickly see that $$ \frac{\ln|2x|}{2} \in \int \frac{1}{2x} dx $$ $$ \frac{\ln|x|}{2} \in \int \frac{1}{2x} dx $$ since their derivatives are both equal to $\frac{1}{2x}$ (derivative of the constant $\frac{\ln|2|}{2}$ "vanishes"). Even though this notation is rarely used, it gives you better insight into what the indefinite integral really is.

Sil
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Note $\frac 12\log|2x| + C_1= \frac 12(\log(2) + \log|x|) + C_1= \frac 12 \log|x| + \frac 12 \log(2) + C_1 = \frac 12\log|x| + C_2$

MathematicsStudent1122
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