\(\begin{align*} \int 2^{lnx}dx \end{align*}\) |
Connaissances:
- Logarithme
- Primitive de \(e^u\)
\(\begin{align*} I & = \int 2^{lnx}dx \\
& = \int \big( e^{ln2}\big)^{lnx} dx\\
& = \int e^{ln2 \times lnx} dx\\
& = \int \big(e^{lnx} \big) ^{ln2} dx\\
& = \int x^{ln2} dx \\
& = \frac{1}{ln2+1}x^{ln2+1} + C\\
& = \frac{1}{ln2+1}x^{ln2}.x^1 + C \\
& = \frac{1}{ln2+1}.2^{lnx}.x +C \\
& = \frac{x. 2^{lnx}}{ln2+1} +C \end{align*}\)
\[\boxed {\begin{align*} I = \frac{x. 2^{lnx}}{ln2+1} +C(\in \mathbb R) \end{align*}}\]