Finding a limit converging into $\ln(x)$ As we know, $\int\dfrac{1}{x}dx = \ln (x) + c$
Also, $\int x^a dx = \dfrac{x^{a+1}}{a+1} + c$ for any $a \neq -1$
But since $x^{-1} = \dfrac{1}{x}$, I suspected that $\ln x = \int\dfrac{1}{x}dx = \int x^{-1} dx = \lim_{a\to -1} \dfrac{x^{a+1}}{a+1}$
However this is false but I then realised this is pretty close from the real equality $\ln x = \lim_{a\to 0} \dfrac{x^a - 1}{a}$
But I don't manage to understand where does this $-1$ come from.
I know what I did was everything but rigourous, but is there a way to get this equality with a similar method to what I've done?
Thank you for reading!

It's probably better to think to
$$
\ln x=\int_{1}^{x} \frac{1}{t}\,dt
$$
Since, for $a\ne -1$,
$$
\int_{1}^{x} t^a\,dt=\frac{x^{a+1}-1}{a+1}
$$
we can conjecture that
$$
\lim_{a\to-1}\frac{x^{a+1}-1}{a+1}=\ln x
$$
which is indeed true, because this is the derivative at $-1$ of the function
$$
f(a)=x^{a+1}
$$
and
$$
f'(a)=x^{a+1}\ln x
$$
When dealing with “indefinite integrals” it's common to get wrong conclusions, because of the constant of integration.
