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Why does $\int \left(-\frac{1}{5x}\right) dx = -\frac{1}{5}\ln(x)+c$ and not $-\frac{1}{5}\ln(5x)+c$ instead?

When I differentiate both solutions I get the same answer, $-\frac{1}{5x}$.

Can anyone give me an explanation as to why? Thank you.

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    $\begingroup$ $\int -\frac{1}{5x} dx=-\frac{1}{5}\int\frac{1}{x}dx$ $\endgroup$ Jul 18, 2020 at 14:48
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    $\begingroup$ The expressions are equivalent since $\ln(x)$ and $\ln(5x)=\ln(5)+\ln(x)$ differ by a constant. $\endgroup$ Jul 18, 2020 at 14:53
  • $\begingroup$ Also note that you are implicitly assuming that the domain is the positive real numbers. You need a different antiderivative for the negative real numbers. $\endgroup$ Jul 18, 2020 at 14:55

5 Answers 5

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Both are correct: on one side, $$ \int -\frac{1}{5x}dx = -\frac15 \int \frac{1}{x}dx = -\frac15\ln(x) + c,$$ because of the linearity of integrals $\int af(x) dx = a\int f(x) dx$ for every constant number $a$ (in particular $a = -\frac15$ in this case).

On the other side, you might also change variable, and set $u = 5x$, if you wish, but in that case $du = 5dx$ and therefore $$ \begin{split} \int - \frac{1}{5x} dx &= - \int \frac{1}{u} \frac{du}{5} \\ &= -\frac{1}{5} \ln(u) + c’ \\ &= -\frac15\ln(5x) + c’ = -\frac15 \ln(x) - \frac15\ln(5) + c’ \end{split} $$ Since $\ln(5x) = \ln(5) + \ln(x)$ it’s just a metter of choosing a different constant $c$ or $c’$.

They clearly both give the same function if you differentiate, for they differ for a constant value $\ln(5)$ whose derivative is zero.

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  • $\begingroup$ Thank you, this really helped. $\endgroup$
    – mikejacob
    Jul 18, 2020 at 15:37
  • $\begingroup$ We need to include the absolute value sign in $\ln|x|$ since we don't know if $x>0$. $\endgroup$
    – Axion004
    Jul 19, 2020 at 3:31
  • $\begingroup$ @Axion004 in which case you need also to use two constants $c$ for $x>0$ and $x<0$ as they could be different. I have assumed $x>0$, albeit implicitly, in order to show why there are two possible solutions which look different but they are the same. $\endgroup$
    – Logos
    Jul 19, 2020 at 9:15
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Well, notice that:

$$\int-\frac{1}{\text{n}x}\space\text{d}x=-\frac{1}{\text{n}}\int\frac{1}{x}\space\text{d}x=\text{C}-\frac{\ln\left|x\right|}{\text{n}}\tag1$$

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  • $\begingroup$ Yeah i get that but is the other way still right? I mean, I got a different answer. $\endgroup$
    – mikejacob
    Jul 18, 2020 at 14:51
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Both answers are equivalent.

Through the logarithm product rule,

$$ -\frac15\ln(5x) + C = -\frac15(\ln(x) + \ln(5)) + C $$

Then simplifying:

$$ \text{LHS} = -\frac15\ln(x) + \left( -\frac15\ln(5) + C \right) $$

Since $C$ is a constant, $-\dfrac15\ln(5) + C$ is a different constant.

Then let $-\dfrac15\ln(5) + C = C_1$. Then

$$ \text{LHS} = -\frac15\ln(x) + C_1 $$

Which is the first expression in your question.

The reason why they both have the same derivative is because they differ by a constant.

Your answer is probably not considered the best answer because it is less “simplified” than the other.

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$\bullet$ See this $$ \int -\frac{dt}{5t} = -\frac{1}{5} \int \frac{dt}{t} = -\frac{1}{5} \ln\lvert t \rvert + C $$

And also this:

$\bullet$ Let $u = 5t \implies du = 5 dt$, and the substitution makes sense as the function is bijective on the whole of $\mathbb{R}$.

therefore, \begin{align*} -\frac{1}{5} \int \frac{du}{u} = -\frac{1}{5} \ln\lvert u \rvert + C = -\frac{1}{5} \ln\lvert 5t \rvert + C \end{align*} Now is it fine @mikejacob ?

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$$\int\frac1x\,dx=\ln\lvert x\rvert+c=\ln\lvert x\rvert+\ln a+c^\prime=\ln\lvert ax\rvert+c^\prime$$

, where $c,c^\prime\in\mathbb R,a\in\mathbb R^+$.

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