\(\begin{align*} \int \sqrt{1+(x-\frac{1}{4x})²}dx \end{align*}\) |
Connaissances:
- calculs simples
Commençons par un changement de variable:
\(\begin{align*} I & = \int \sqrt{1+(x-\frac{1}{4x})²}dx \\
& = \int \sqrt{1+(x²-\frac{1}{2}+\frac{1}{16x²})}dx \\
& = \int \sqrt{x²+\frac{1}{2}+\frac{1}{16x²}}dx \\
& = \int \sqrt{(x+\frac{1}{4x})²}dx \\
& = \int (x+\frac{1}{4x})dx \\
& = \frac{1}{2}x²+\frac{1}{4}ln \lvert x \rvert +C \\ \end{align*}\)
\[\boxed {\begin{align*} I = \frac{1}{2}x²+\frac{1}{4}ln \lvert x \rvert +C(\in \mathbb R) \end{align*}}\]