- #1
protonchain
- 98
- 0
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]
Last edited: