Let y = [tex] 2^{lnx} [/tex]
lnx = log₂y
lnx = [tex] \frac{lny}{ln2} [/tex]
lnx·ln2 = lny
y = [tex] e^{lnxln2} [/tex]
y = [tex] x^{ln2}[/tex]
∴ the integral of x^(㏑2) is [tex] \frac{x^(ln2+1)}{ln2 + 1} [/tex]
So, with the bounds, it becomes:
[tex] \frac{1}{ln2 + 1} [/tex]
