Integration of $\frac{f'(x)}{f(x)}$? $\frac{f'(x)}{f(x)}$ 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?
 A: $$
1+\int f^{-1}(x)dx = 1+\ln\rvert f(x)\rvert+c
$$
Wait!
$$
\int \frac{f'(x)}{f(x)} dx = \ln|f(x)| + c$$
but
$$
\int \frac1{f(x)} dx \ne\ln|f(x)| + c
$$
A: There were two errors:
Error 1:

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


Error 2:

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

A: 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.
A: $C$ is just shorthand for "up to some constant additive value", so that your $C+1$ can be written as just $C$.
A: 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 $C$.  $$$$An additional point: Never write $f^{-1}(x)$ for $\dfrac{1}{f(x)}$. $f^{-1}(x)$ is the inverse function of $f(x)$ ie if you plug in a value $x=a$ into $f(x)$, obtain an output value and plug it into $f^{-1}(x)$, then you will get the same $a$ that you started with.
A: Yes, you are correct. In fact you can always summerize constants.
Another example:
Let's take two approaches to finding
for $x\in (-1,1)$ $$\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.
A: Any function of arbitrary constants is also another arbitrary constant.
