NEW FUNCTION

Function Expression :

\[f(x)=\frac{x+1}{x^2+2x} \]

Domain

\[\left]-\infty, -2\right[ \cup \left]-2, 0\right[ \cup \left]0, \infty\right[ \]

Limits

\[\lim_{x \rightarrow-\infty}f(x) = 0 \]
\[\lim_{x \overset{<}{\rightarrow-2} }f(x) = -\infty \]
\[\lim_{x \overset{>}{\rightarrow-2} }f(x) = +\infty \]
\[\lim_{x \overset{<}{\rightarrow0} }f(x) = -\infty \]
\[\lim_{x \overset{>}{\rightarrow0} }f(x) = +\infty \]
\[\lim_{x \rightarrow+\infty}f(x) = 0 \]
\[ \]

Derivate

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

Integral

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

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