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

Question 1

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 $\int \frac{1}{x} \,\mathrm dx$ :

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

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.

Question 2

If you want to try to prove $\int\frac{\mathrm dx}x=\ln x + C$ (for $x \gt 0$ ), try the substitution

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

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 $x=e^u,u=\ln x$ .

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

Answer

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