Question Number 168898 by mathlove last updated on 20/Apr/22
Answered by FelipeLz last updated on 21/Apr/22
![2^x = u ⇒ du = 2^x ln(2)dx ∫2^2^2^x 2^2^x 2^x dx = (1/(ln(2)))∫2^2^u 2^u du 2^u = t ⇒ dt = 2^u ln(2)du (1/(ln(2)))∫2^2^u 2^u du = (1/([ln(2)]^2 ))∫2^t dt = (2^t /([ln(2)]^3 ))+c = (2^2^(2x) /([ln(2)]^3 ))+c](https://www.tinkutara.com/question/Q168904.png)
$$\mathrm{2}^{{x}} \:=\:{u}\:\Rightarrow\:{du}\:=\:\mathrm{2}^{{x}} \mathrm{ln}\left(\mathrm{2}\right){dx} \\ $$$$\int\mathrm{2}^{\mathrm{2}^{\mathrm{2}^{{x}} } } \mathrm{2}^{\mathrm{2}^{{x}} } \mathrm{2}^{{x}} {dx}\:=\:\frac{\mathrm{1}}{\mathrm{ln}\left(\mathrm{2}\right)}\int\mathrm{2}^{\mathrm{2}^{{u}} } \mathrm{2}^{{u}} {du} \\ $$$$\mathrm{2}^{{u}} \:=\:{t}\:\Rightarrow\:{dt}\:=\:\mathrm{2}^{{u}} \mathrm{ln}\left(\mathrm{2}\right){du} \\ $$$$\frac{\mathrm{1}}{\mathrm{ln}\left(\mathrm{2}\right)}\int\mathrm{2}^{\mathrm{2}^{{u}} } \mathrm{2}^{{u}} {du}\:=\:\frac{\mathrm{1}}{\left[\mathrm{ln}\left(\mathrm{2}\right)\right]^{\mathrm{2}} }\int\mathrm{2}^{{t}} {dt}\:=\:\frac{\mathrm{2}^{{t}} }{\left[\mathrm{ln}\left(\mathrm{2}\right)\right]^{\mathrm{3}} }+{c}\:=\:\frac{\mathrm{2}^{\mathrm{2}^{\mathrm{2}{x}} } }{\left[\mathrm{ln}\left(\mathrm{2}\right)\right]^{\mathrm{3}} }+{c}\:\:\: \\ $$