Demystify integration of $\int \frac{1}{x} \mathrm dx$ I've learned in my analysis class, that


  $$ \int \frac{1}{x} \mathrm dx = \ln(x). $$


I can live with that, and it's what I use when solving equations like that.
But how can I solve this, without knowing that beforehand.

Assuming the standard rule for integration is


  $$ \int x^a \, \mathrm dx = \frac{1}{a+1} \cdot x^{a+1} + C .$$


If I use that and apply this to $\int \frac{1}{x} \,\mathrm dx$:

$$ \begin{align*}
  \int \frac{1}{x}\mathrm  dx &amp;= \int x^{-1} \,\mathrm dx \\
                      &amp;= \frac{1}{-1+1} \cdot x^{-1+1} \\
                      &amp;= \frac{x^0}{0}
\end{align*} $$

Obviously, this doesn't work, as I get a division by $0$. I don't really see, how I can end up with $\ln(x)$. There seems to be something very fundamental that I'm missing.

I study computer sciences, so, we usually omit things like in-depth math theory like that. We just learned that $\int \frac{1}{x} dx = \ln(x)$ and that's what we use.
If you want to try to prove $\int\frac{\mathrm dx}x=\ln x + C $ (for $x \gt 0$), try the substitution

$$ \begin{align}
  x &amp;= e^u \\
  \mathrm dx &amp;= e^u \mathrm du
\end{align} $$

This substitution is justified because the exponential function is bijective from $\mathbb{R}$ to $(0,\infty)$ (hence for every $x$ there exists a $u$) and continuously differentiable (which allows an integration by substitution).

$$\int\frac{\mathrm dx}x=\int\frac{e^u\mathrm du}{e^u}=u+C$$

Now just use the fact that natural log is the inverse of the exponential function.  If $x=e^u,u=\ln x$.
I did sign up just to say some answers here are misleading.

$$\int \frac{1}{x} dx = \ln(x). $$

is not  a calculation but a definition. This is how ln(x) is defined. ln(x) is defined before exp(x)

http://en.wikipedia.org/wiki/Natural_logarithm#Definitions