NEW FUNCTION

Function Expression :

\[f(x)=ln(\frac{1+x}{1-x} )-\frac{1}{x} \]

Domain

\[\left]-1, 0\right[ \cup \left]0, 1\right[ \]

Limits

\[\lim_{x \overset{>}{\rightarrow-1} }f(x) = -\infty \]
\[\lim_{x \overset{<}{\rightarrow0} }f(x) = +\infty \]
\[\lim_{x \overset{>}{\rightarrow0} }f(x) = -\infty \]
\[\lim_{x \overset{<}{\rightarrow1} }f(x) = +\infty \]
\[ \]

Derivate

\[f^{\,\prime}(x)=\frac{\left(1 - x\right) \left(\frac{1}{1 - x} + \frac{x + 1}{\left(1 - x\right)^{2}}\right)}{x + 1} + \frac{1}{x^{2}} \]
\[f^{\,\prime}(x)=\frac{- x^{2} - 1}{x^{4} - x^{2}} \]
\[f^{\,\prime}(x)=\frac{- x^{2} - 1}{x^{2} \left(x^{2} - 1\right)} \]

Integral

\[F(x) = x \log{\left(- \frac{x}{x - 1} - \frac{1}{x - 1} \right)} - \log{\left(x \right)} + 2 \log{\left(x + 1 \right)} - \log{\left(- \frac{x}{x - 1} - \frac{1}{x - 1} \right)} \]

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