# Thread: Help proving an integration formula via differentiation

1. ## Help proving an integration formula via differentiation

I'm stuck on something simple, I'd be grateful for any help.

My text ("Forgotten Calculus", Bleau) states the following integration formula:

integral [dx / (x ln x)] = ln |ln x| + C.

In the course of a problem where I've applied the formula, I realised I can't prove it.

Here's my attempt, for what it's worth.

(Re notation:
exponents are denoted with the "^" sign, e.g x squared = x^2;
first derivatives are denoted with the " ' " symbol, e.g. the 1st derivative of f(x) = f'(x))

f(x) = ln |ln x| + C

I want to differentiate this via the product rule, so I rewrite the function as

f(x) = (ln x^0) * (ln x) + C

Using the product rule,

f'(x) = [ (ln x^0) * (ln x)' ] + [(ln x^0)' * (ln x) ]
= [ (ln 1) * (1/x)] + [ (0/1) * (ln x)]
= [ ln / x ] + [0]
= ln / x.

I don't see how ln / x = ln |ln x|.

2. Originally Posted by lingyai
I'm stuck on something simple, I'd be grateful for any help.

My text ("Forgotten Calculus", Bleau) states the following integration formula:

integral [dx / (x ln x)] = ln |ln x| + C.

In the course of a problem where I've applied the formula, I realised I can't prove it.

Here's my attempt, for what it's worth.

(Re notation:
exponents are denoted with the "^" sign, e.g x squared = x^2;
first derivatives are denoted with the " ' " symbol, e.g. the 1st derivative of f(x) = f'(x))

f(x) = ln |ln x| + C

I want to differentiate this via the product rule, so I rewrite the function as

f(x) = (ln x^0) * (ln x) + C
$\ln |\ln(x)| =\ln(|\ln(x)|)$,

you need to use the chain rule.

RonL

3. You have, $x>0$
$\int \frac{dx}{x\ln x}$
Let, $u=\ln x$ then,
$du/dx=1/x$
Thus,
$\int \frac{1}{\ln x} \cdot \frac{1}{x} dx$
Gives after substitution,
$\int \frac{1}{u} \frac{du}{dx} dx=\int \frac{1}{u} du$
Thus,
$\ln|u|+C$
Thus, substitute back,
$\ln|\ln x|+C$