Can you have many integrations for a single function? (sorry if the terminology is wrong)

Forgive me for the amateur question (first time here), but it's bugging me!

OK, for example, integrating $\dfrac{1}{5x}$ would yield what? $\dfrac{\ln(5x)}{5} +C$ OR $\dfrac{\ln(x)}{5} +C$ ? Both of these when differentiated form $\dfrac{1}{5x}$.

And if you say that you must take the constant out, why shouldn't you do the same here then? integration of $\dfrac{1}{(5x+3)} = \dfrac{1}{5} \ln(5x+3)+C$ rather than $ \dfrac{1}{5} \ln(x+\dfrac35)+C$ And for both examples, each integral is fundamentally different, one cannot be simplified to reach the other one. Check by substituting values of $x$ into both.

If somehow both are right, which do you choose if you are asked to write it out in terms of $x$ and $C$. (not defining the constant of integration).

On Wikipedia it does state that the constant of integration can be different and so you can have many 'anti derivatives' for a function (not really sure what that means), but in an A level text book, they just have one answer with 'C'. So does this mean $\dfrac{1}{5x}$ and $\dfrac{1}{(5x+3)}$ have two (or more) anti derivatives?

So my question really becomes, how do you reach the simplest anti derivative possible, if that is what is required by the textbook.

Thank you for your time sir!

Solution 1:

The two functions you mention actually differ only by a constant. Note that $\ln(5x)/5+C=\ln(x)/5+(\ln(5)/5+C)=\ln(x)/5+D$. They both have the same derivative and are both anti derivatives of $1/5x$.

If you choose initial conditions, say demanding that the anti derivative satisfies $F(1/5)=0$, and you work out the correct constant, then you will get the same result despite the different appearance.

Solution 2:

All integrands of a function differ by a constant, i.e. if you have $f_1,f_2$ such that $f_1' = f_2' = f$, then $f_1=f_2 + C$ for some constant $C$.

In your examples:

The first example: $$f_1(x) = \frac15 \ln(5x) = \frac{1}{5}(\ln 5 + \ln x) = \frac15\ln 5 + \frac15\ln x=\frac{1}{5}\ln 5 + f_2(x),$$ meaning that the two functions differ only by a constant, that constant being $\frac{\ln 5}{5}$.

The second example:

$$\frac{1}{5}\ln{(x+\frac35)} = \frac15 \ln(\frac15(5x+3)) = \frac15\cdot \ln\frac15 + \frac15\cdot \ln (3x+5)$$ so again, the two functions differ by a constant.