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:
$$
\begin{align*}
&amp;\int\frac{\frac{d}{dx}f(x)}{f(x)}dx \\
=\quad&amp;\int\frac{d}{dx}(f(x))\cdot (f(x))^{-1}dx \\
=\quad&amp;f(x)\cdot (f(x))^{-1}-\int -f(x)\cdot (f(x))^{-2}dx \\
=\quad&amp;1+\int (f(x))^{-1}dx \\
=\quad&amp;1+\ln\rvert f(x)\rvert+c
\end{align*}
$$
Is there a mistake or can I summarize $1+c$, because $1$ is a constant?
$$
1+\int f^{-1}(x)dx = 1+\ln\rvert f(x)\rvert+c
$$
Wait!
$$
\int \frac{f'(x)}{f(x)} dx = \ln|f(x)| + c$$
but
$$
\int \frac1{f(x)} dx \ne\ln|f(x)| + c
$$
Any function of arbitrary constants is also another arbitrary constant.