\(\begin{align*} \int \frac{x-1}{x^4-1}dx \end{align*}\) |
Connaissances:
- Décomposition en éléments simples
\(\begin{align*} \frac{x-1}{x^4-1} & = \frac{x-1}{(x²-1)(x²+1)} \\
& = \frac{x-1}{(x-1)(x+1)(x²+1)} \\
& = \frac{1}{(x+1)(x²+1)} = \frac{\overbrace{a}^{1/2}}{x+1}+ \frac{bx+\overbrace{c}^{-1/2}}{x²+1} \end{align*}\)
\(\begin{align*} \begin{cases} \times (x+1) \text{ et } x=-1 & \Rightarrow & a=1/2 \\
x=0 \Rightarrow 1 = a+c & \Rightarrow & c=1/2 \\
x= 1 \Rightarrow 1/4 = 1/4 +\frac{b +1/2}{2} & \Rightarrow &b=-1/2
\end{cases} \end{align*}\)
\(\begin{align*} I & = \int \frac{x-1}{x^4-1}dx \\
& =\int (\frac{1}{2}\frac{1}{x+1}-\frac{1}{2}\frac{x}{x²+1}+\frac{1}{2}\frac{1}{x²+1})dx \\
& =\frac{1}{2}\int \frac{1}{x+1}dx -\frac{1}{4} \int \frac{2x}{x²+1}dx+\frac{1}{2} \int \frac{1}{x²+1}dx \\
& = \frac{1}{2}ln \lvert x+1 \rvert - \frac{1}{4}ln \lvert x²+1 \rvert + \frac{1}{2}tan^{-1}x +C \\
& = \frac{1}{2}ln \lvert x+1 \rvert - \frac{1}{4}ln (x²+1) + \frac{1}{2}tan^{-1}x +C
\end{align*}\)
\[\boxed {\begin{align*} I =\frac{1}{2}ln \lvert x+1 \rvert - \frac{1}{4}ln (x²+1) + \frac{1}{2}tan^{-1}x +C(\in \mathbb R) \end{align*}}\]