How to integrate 1/x * 1/ln(x) dx by parts: I get: ∫1⁄x 1⁄ln(x) dx = ln(x) * 1/ln(x) - ∫ ln(x) (-ln^-2(x) * 1/x) dx = 1 + ∫ 1/ln(x) * 1/x dx => 0=1

[usk]

How to integrate 1/x * 1/ln(x) dx by parts

I get:
∫1⁄x 1⁄ln(x) dx = ln(x) * 1/ln(x) - ∫ ln(x) (-ln^-2(x) * 1/x) dx = 1 + ∫ 1/ln(x) * 1/x dx
=> 0=1 which is absurd

 

[BigBeachBanana]

I get:
∫1⁄x 1⁄ln(x) dx = ln(x) * 1/ln(x) - ∫ ln(x) (-ln^-2(x) * 1/x) dx = 1 + ∫ 1/ln(x) * 1/x dx
=> 0=1 which is absurd

There's nothing wrong with this so far.

[math]\int \dfrac{1}{x\ln x} \, dx = 1 + \int \dfrac{1}{x\ln x} \, dx[/math]
The issue is what conclusion you're making.
You're assuming that
[math]\int f(x) \, dx - \int f(x)\, dx=0[/math]
But what it means is
[math]\int f(x) \, dx - \int f(x) \, dx = \int f(x) - f(x) \, dx = \int 0 \, dx= C[/math]
This only shows that [imath]C=1[/imath] is a constant.

PS: The integral can be solved by u-sub. Hint: [math]\dfrac{d}{dx}\left( \ln x \right) = \dfrac{1}{x}[/math]