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Question Number 43550 by peter frank last updated on 11/Sep/18

$$provethat$$$${\int }_{}{4}_{}^{4^{4^x}}.4^{4^x}.4^{x}dx=\frac{4^{4^x}}{\left(\log \underset{e}{4}\right)}$$

Commented by peter frank last updated on 15/Sep/18

$$helpplz$$

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