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Question Number 168898 by mathlove last updated on 20/Apr/22

Answered by FelipeLz last updated on 21/Apr/22

$$2^x=u\Rightarrow du=2^x\ln \left(2\right)dx$$$$\int 2^{2^{2^x}}{2}^{2^x}{2}^{x}dx=\frac{1}{\ln \left(2\right)}\int 2^{2^u}{2}^{u}du$$$$2^u=t\Rightarrow dt=2^u\ln \left(2\right)du$$$$\frac{1}{\ln \left(2\right)}\int 2^{2^u}{2}^{u}du=\frac{1}{{\left[\ln \left(2\right)\right]}^2}\int 2^{t}dt=\frac{2^t}{{\left[\ln \left(2\right)\right]}^3}+c=\frac{2^{2^{2x}}}{{\left[\ln \left(2\right)\right]}^3}+c$$

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