\(\begin{align*} \int \frac{1}{x^3-4x²}.dx \end{align*}\) |
Connaissances:
- Décomposition en éléments simples
\(\begin{align*} I & = \int \frac{1}{x^3-4x²}.dx \\
& = \int \frac{1}{x²(x-4)}dx \\
& = \int (\frac{ax+b}{x²} + \frac{c}{x-4})dx \\
& = \int (\frac{ax}{x²} + \frac{b}{x²} + \frac{c}{x-4})dx \\
& = \int (\frac{a}{x} + \frac{b}{x²} + \frac{c}{x-4})dx \\ \\
& \begin{cases} \times (x-4) \text{ et }x=4 \Rightarrow & c= 1/16 \\
\times x² \text{ et } x=0 \Rightarrow & b = -1/4 \\
x=1 \Rightarrow -1/3 = a-1/4-1/48 \Rightarrow & a=-3/48=-1/16
\end{cases} \\ \\
I & = \int \bigg(\frac{-1/16}{x} + \frac{-1/4}{x²} + \frac{1/16}{x-4} \bigg)dx \\
& = -\frac{1}{16}\int \frac{dx}{x} -\frac{1}{4} \int \frac{dx}{x²} + \frac{1}{16} \int \frac{dx}{x-4} \\
& = -\frac{1}{16}ln \lvert x \rvert )+\frac{1}{4}\frac{1}{x} +\frac{1}{16}ln \lvert x-4 \rvert+C \\
& = \frac{1}{16} ln \lvert \frac{x-4}{x} \rvert +\frac{1}{4x}+ C
\end{align*}\)
\[\boxed {\begin{align*} I = \frac{1}{16} ln \bigg( \lvert \frac{x-4}{x} \rvert \bigg) +\frac{1}{4x}+ C(\in \mathbb R) \end{align*}}\]