NEW FUNCTION

Function Expression :

\[f(x)=\frac{2x+3}{x^2(x+1 )} \]

Domain

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

Limits

\[\lim_{x \rightarrow-\infty}f(x) = 0 \]
\[\lim_{x \overset{<}{\rightarrow-1} }f(x) = -\infty \]
\[\lim_{x \overset{>}{\rightarrow-1} }f(x) = +\infty \]
\[\lim_{x \overset{<}{\rightarrow0} }f(x) = +\infty \]
\[\lim_{x \overset{>}{\rightarrow0} }f(x) = +\infty \]
\[\lim_{x \rightarrow+\infty}f(x) = 0 \]
\[ \]

Derivate

\[f^{\,\prime}(x)=\frac{2}{x^{2} \left(x + 1\right)} + \frac{\left(2 x + 3\right) \left(- x^{2} - 2 x \left(x + 1\right)\right)}{x^{4} \left(x + 1\right)^{2}} \]
\[f^{\,\prime}(x)=\frac{- 4 x^{2} - 11 x - 6}{x^{3} \left(x^{2} + 2 x + 1\right)} \]
\[ \]

Integral

\[F(x) = - \log{\left(x \right)} + \log{\left(x + 1 \right)} - \frac{3}{x} \]

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