Ikunupe6v

2021-12-31

Demystify integration of $\int \frac{1}{x}dx$
Ive

Marcus Herman

If you want to try to prove , try the substitution
$x={e}^{u}$
$dx={e}^{u}du$
This substitution is justified because the exponential function is bijective from $\mathbb{R}$ to $\left(0,\mathrm{\infty }\right)$ (hence for every x there exists a u) and continuously differentiable (which allows an integration by substitution).
$\int \frac{dx}{x}=\int \frac{{e}^{u}du}{{e}^{u}}=u+C$
Now just use the fact that natural log is the inverse of the exponential function. If
$x={e}^{u},u=\mathrm{ln}x$

Ronnie Schechter

Lets

Vasquez

To show $\int \frac{dx}{x}=\mathrm{ln}x+C$ for positive x, we can show that the derivative of $\mathrm{ln}x$ is $\frac{1}{x}$. This can be done using the standard limit
$\underset{x\to 0}{lim}\frac{\mathrm{ln}\left(1+x\right)}{x}=1$
and the definition of the derivative. We have
$\frac{d}{dx}\left(\mathrm{ln}x\right)=\underset{h\to 0}{lim}\frac{\mathrm{ln}\left(x+h\right)-\mathrm{ln}x}{h}=\underset{h\to 0}{lim}\frac{1}{h}\mathrm{ln}\left(1+\frac{h}{x}\right)=\left[t=\frac{1}{x}\right]=\underset{t\to 0}{lim}\frac{1}{x}$
$\cdot \frac{\mathrm{ln}\left(1+t\right)}{t}=\frac{1}{x}$
which is what we wanted to show.

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