- #1

- 96

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## Main Question or Discussion Point

Another one of those 0 = 1 proofs.

If you dislike them, please navigate away to cats doing things with captions. Otherwise stay tuned.

Tell me what's wrong (obviously I know what's wrong but I just want to put it out there for people to mull over).

Note, you need to know basic calculus.

[tex]

\int \frac{1}{x} dx = \int \frac{1}{x} dx

[/tex]

[tex]

u = \frac{1}{x}

[/tex]

[tex]

dv = dx

[/tex]

[tex]

du = \frac{-1}{x^2} dx

[/tex]

[tex]

v = x

[/tex]

[tex]

\int \frac{1}{x} dx = u * v - \int v du

[/tex]

[tex]

\int \frac{1}{x} dx = \frac{1}{x} * x - \int x * \frac{-1}{x^2} dx

[/tex]

[tex]

\int \frac{1}{x} dx = 1 - \int \frac{-1}{x} dx

[/tex]

[tex]

\int \frac{1}{x} dx = 1 + \int \frac{1}{x} dx

[/tex]

[tex]

0 = 1

[/tex]

If you dislike them, please navigate away to cats doing things with captions. Otherwise stay tuned.

Tell me what's wrong (obviously I know what's wrong but I just want to put it out there for people to mull over).

Note, you need to know basic calculus.

[tex]

\int \frac{1}{x} dx = \int \frac{1}{x} dx

[/tex]

[tex]

u = \frac{1}{x}

[/tex]

[tex]

dv = dx

[/tex]

[tex]

du = \frac{-1}{x^2} dx

[/tex]

[tex]

v = x

[/tex]

[tex]

\int \frac{1}{x} dx = u * v - \int v du

[/tex]

[tex]

\int \frac{1}{x} dx = \frac{1}{x} * x - \int x * \frac{-1}{x^2} dx

[/tex]

[tex]

\int \frac{1}{x} dx = 1 - \int \frac{-1}{x} dx

[/tex]

[tex]

\int \frac{1}{x} dx = 1 + \int \frac{1}{x} dx

[/tex]

[tex]

0 = 1

[/tex]

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