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

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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
 
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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.

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

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