ChaoticLlama
Junior Member
- Joined
- Dec 11, 2004
- Messages
- 199
Consider the function \(\displaystyle \L\ y = \arctan \left( {\frac{1}{x}} \right)\).
To the best of my knowledge;
\(\displaystyle \L\begin{array}{c}
\frac{{dy}}{{dx}} = \frac{1}{{1 + \frac{1}{{x^2 }}}}\left( {\frac{{ - 1}}{{x^2 }}} \right) \\
= \frac{{ - 1}}{{x^2 + 1}} \\
\end{array}\)
and if you integrate the new function, you get;
\(\displaystyle \L\int {\frac{{ - dx}}{{x^2 + 1}}} = - \arctan (x) + C\)
Now unless I'm missing something here; I think I broke Calculus.
To the best of my knowledge;
\(\displaystyle \L\begin{array}{c}
\frac{{dy}}{{dx}} = \frac{1}{{1 + \frac{1}{{x^2 }}}}\left( {\frac{{ - 1}}{{x^2 }}} \right) \\
= \frac{{ - 1}}{{x^2 + 1}} \\
\end{array}\)
and if you integrate the new function, you get;
\(\displaystyle \L\int {\frac{{ - dx}}{{x^2 + 1}}} = - \arctan (x) + C\)
Now unless I'm missing something here; I think I broke Calculus.