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?

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.
