Demystify integration of \int \frac{1}{x} \mathrm dx\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:

\int \frac{1}{x}\mathrm dx &= \int x^{-1} \,\mathrm dx \\
&= \frac{1}{-1+1} \cdot x^{-1+1} \\
&= \frac{x^0}{0}

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

x &= e^u \\
\mathrm dx &= e^u \mathrm du

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.

Source : Link , Question Author : polemon , Answer Author : Ali Caglayan

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