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\( \begin{align*} I & = \int \frac{1}{x^4+x}dx \\
& = \int \frac{1}{x(x^3+1)}dx \\
& = \int \frac{1}{x(x+1)(x²-x+1)}dx \\
\end{align*}\)
Procédons maintenant à une décomposition en éléments simples:
\( \begin{align*}& \frac{1}{x(x+1)(x²-x+1)} = \frac{A}{x} +\frac{B}{x+1} + \frac{Cx+D}{x²-x+1} \\
& \begin{cases} \times (x+1) \text{ et } x=-1 \Rightarrow \frac{1}{-1((-1)²-1-1)}=B & \Rightarrow B=1 \\
\times x \text{ et }x=0 \Rightarrow \frac{1}{(0+1)(0²-0+1)}=A & \Rightarrow A=1 \\
\Rightarrow \frac{1}{x(x+1)(x²-x+1)} = \frac{1}{x} +\frac{1}{x+1} + \frac{Cx+D}{x²-x+1} \\
x=1 \Rightarrow \frac{1}{2}= 1+\frac{1}{2} + \frac{C+D}{1} & \Rightarrow C+D=-1 \\
x=2 \Rightarrow \frac{1}{2(2+1)(2²-2+1}= \frac{1}{2}+\frac{1}{2+1}+ \frac{2C+D}{2²-2+1} \\
\Rightarrow \frac{1}{18}= \frac{5}{6}+ \frac{2C+D}{3} & \Rightarrow 2C+D = \frac{-14}{6}=\frac{-7}{3} \\
\Rightarrow \begin{cases}C=\frac{-7}{3}+1 = \frac{-4}{3} & & \Rightarrow C= \frac{-4}{3} \\
D= -1-C=-1+ \frac{4}{3} & & \Rightarrow D = \frac{1}{3}
\end{cases} \end{cases} \\
&\Rightarrow \frac{1}{x(x+1)(x²-x+1)} = \frac{1}{x} +\frac{1}{x+1} + \frac{-4/3x+1/3}{x²-x+1}
\end{align*}\)
Autre méthode :
\( \begin{align*} I & = \int \frac{1}{x^4+x}dx \\
& = \int \frac{1}{x^4(1+x^{-3})}dx \\
& = \int \frac{x^{-4}}{1+x^{-3}}dx \\
\end{align*}\)
On remarque une forme en \(u'/u\). Procédons maintenant à un changement de variable:
\( \begin{align*} & \begin{cases} u = 1+x^{-3} \\ du = -3x^{-4}dx \Rightarrow dx =\frac{du}{-3x^{-4}} \end{cases} \\
I & = \int \frac{x^{-4}}{1+x^{-3}}dx \\
& = \int \frac{x^{-4}}{u} \times \frac{du}{-3x^{-4}} \\
& = -\frac{1}{3}\int \frac{du}{u} \\
& = -\frac{1}{3} ln \lvert u \rvert +C \\
& = -\frac{1}{3} ln \lvert 1+x^{-3} \rvert +C
\end{align*}\)
\[ \boxed{ \begin{align*} I = -\frac{1}{3} ln \lvert 1+x^{-3} \rvert +C(\in \mathbb R) \end{align*}}\]
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