Answered: "Demystify integration of ∫1xdx Ive"

Ikunupe6v

Answered question

2021-12-31

Demystify integration of

\int \frac{1}{x} d x

Ive

Answer & Explanation

Marcus Herman

Beginner2022-01-01Added 41 answers

If you want to try to prove

\int \frac{d x}{x} = \ln x + C (for >0)

, try the substitution

x = e^{u}

d x = e^{u} d u

This substitution is justified because the exponential function is bijective from

R

(0, \infty)

(hence for every x there exists a u) and continuously differentiable (which allows an integration by substitution).

\int \frac{d x}{x} = \int \frac{e^{u} d u}{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

Ronnie Schechter

Beginner2022-01-02Added 27 answers

Lets

Vasquez

Expert2022-01-09Added 669 answers

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

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