Integration of $\frac{f'(x)}{f(x)}$?

Question

$\frac{f'(x)}{f(x)}$ integrated must be $$\int\frac{f'(x)}{f(x)}dx=\ln\rvert f(x)\rvert+c.$$ But when trying to show this with partial integration I get another result: $$ \begin{align*} &\int\frac{\frac{d}{dx}f(x)}{f(x)}dx \\ =\quad&\int\frac{d}{dx}(f(x))\cdot (f(x))^{-1}dx \\ =\quad&f(x)\cdot (f(x))^{-1}-\int -f(x)\cdot (f(x))^{-2}dx \\ =\quad&1+\int (f(x))^{-1}dx \\ =\quad&1+\ln\rvert f(x)\rvert+c \end{align*} $$ Is there a mistake or can I summarize $1+c$, because $1$ is a constant?

Answer 1

$$ 1+\int f^{-1}(x)dx = 1+\ln\rvert f(x)\rvert+c $$ Wait! $$ \int \frac{f'(x)}{f(x)} dx = \ln|f(x)| + c$$ but $$ \int \frac1{f(x)} dx \ne\ln|f(x)| + c $$

Answer 2

There were two errors:

Error 1:

$$\begin{align} \int\frac{df(x)}{dx}(f(x))^{-1}dx &=f(x) (f(x))^{-1}+\int f(x)(f(x))^{-2}\frac{df(x)}{dx}dx \\\\ &\ne f(x) (f(x))^{-1}-\int -f(x) (f(x))^{-2}dx \\ \end{align}$$

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Error 2:

$$\begin{align} 1+\int (f(x))^{-1}dx &\ne 1+\ln\rvert f(x)\rvert+C \end{align}$$

Answer 3

Since you never specify what $c$ was in the first place, it doesn't matter if you have $1+c$ or $c$ as an integration constant. You can always choose a specific $c$ at a later point in time if you are given some initial condition. So your partial integration result is consistent with the first integral you calculated. For example, if you were given $$g(x) = \int_0^x\frac{f'(t)}{f(t)}dt $$ and told that $g(0) = 2$, $f(0) = 1$ you'd know that $g(1) =\ln\rvert f(0)\rvert+c = 0+c = 2$ So you'd take $c = 2$. If you choose to use $g(x) = 1+\left|f(x)\right|+c$ then you'd end up with $c = 1$. You can safely go about this problem in either way.

Answer 4

$C$ is just shorthand for "up to some constant additive value", so that your $C+1$ can be written as just $C$.

Answer 5

This is one of the things in Integration. We just 'gobble up' all the individual constants we get as a result of Integrations and express them as $1$ single, final constant $C$. $$$$An additional point: Never write $f^{-1}(x)$ for $\dfrac{1}{f(x)}$. $f^{-1}(x)$ is the inverse function of $f(x)$ ie if you plug in a value $x=a$ into $f(x)$, obtain an output value and plug it into $f^{-1}(x)$, then you will get the same $a$ that you started with.

Answer 6

Yes, you are correct. In fact you can always summerize constants.
Another example:
Let's take two approaches to finding for $x\in (-1,1)$ $$\int -\frac{1}{\sqrt {1-x^2}}dx$$ Approach 1:
Since $$(\arccos x)'=-\frac{1}{\sqrt{1-x^2}}$$ we have $$\int -\frac{1}{\sqrt {1-x^2}}dx=\arccos x+C \tag{1}$$

Approach 2:
Since $$(\arcsin x)'=\frac{1}{\sqrt{1-x^2}}$$ We have $$\int -\frac{1}{\sqrt {1-x^2}}dx=-\arcsin x +C$$
But $$\arccos x+\arcsin x=\frac{\pi}{2}$$ Therefore $$\int -\frac{1}{\sqrt {1-x^2}}dx=\arccos x-\frac{\pi}{2} +C\tag{2}$$ They are equivalent.

Answer 7

Any function of arbitrary constants is also another arbitrary constant.