\(\begin{align*} \int \frac{1}{\sqrt[3]x+1}dx \end{align*}\) |
Connaissances:
- Changement de variables
\(\begin{align*} I & = \int \frac{1}{\sqrt[3]x+1}.dx \\ \end{align*}\)
Faisons le changement de variable:
\(\begin{align*} u = \sqrt[3]x+1 & \Rightarrow x = (u-1)^3 \\
& \Rightarrow dx = 3(u-1)²du \\ \end{align*}\)
\(\begin{align*} I & = \int \frac{1}{u}.3.(u-1)²du \\
& = 3\int \frac{1}{u}(u²-2u+1)²du \\
& = 3\int (u-2+\frac{1}{u})du \\
& = 3(\frac{1}{2}u²-2u+ln \lvert u \rvert +C \\
& = \frac{3}{2}(\sqrt[3]x+1)²-6(\sqrt[3]x+1) +3ln \lvert \sqrt[3]x+1 \rvert +C
\end{align*}\)
\[\boxed {\begin{align*} I = \frac{3}{2}(\sqrt[3]x+1)²-6(\sqrt[3]x+1) +3ln \lvert \sqrt[3]x+1 \rvert +C(\in \mathbb R) \end{align*}}\]