If c is any arbitrary constant,then int {{2^{ | vedclass

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(B) Let $I = \int {{2^{{2^{{2^x}}}}}{2^{{2^x}}}{2^x}dx}$.Substitute $u = 2^x$,then $du = 2^x \ln 2 \, dx$,so $2^x \, dx = \frac{du}{\ln 2}$.Now,let $v

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$\frac{{{2^{{2^{{2^x}}}}}}}{{{{(\ln 2)}^3}}} + c$

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(B) Let $I = \int {{2^{{2^{{2^x}}}}}{2^{{2^x}}}{2^x}dx}$.Substitute $u = 2^x$,then $du = 2^x \ln 2 \, dx$,so $2^x \, dx = \frac{du}{\ln 2}$.Now,let $v = 2^u = 2^{2^x}$,then $dv = 2^u \ln 2 \, du = 2^{2^x} \ln 2 \, du$,so $2^{2^x} \, du = \frac{dv}{\ln 2}$.Finally,let $t = 2^v = 2^{2^{2^x}}$,then $dt = 2^v \ln 2 \, dv = 2^{2^{2^x}} \ln 2 \, dv$,so $2^{2^{2^x}} \, dv = \frac{dt}{\ln 2}$.Alternatively,using substitution $t = 2^{2^{2^x}}$,we differentiate with respect to $x$:$\frac{dt}{dx} = 2^{2^{2^x}} \cdot \ln 2 \cdot \frac{d}{dx}(2^{2^x}) = 2^{2^{2^x}} \cdot \ln 2 \cdot (2^{2^x} \cdot \ln 2 \cdot \frac{d}{dx}(2^x)) = 2^{2^{2^x}} \cdot 2^{2^x} \cdot 2^x \cdot (\ln 2)^3$.Thus,$2^{2^{2^x}} \cdot 2^{2^x} \cdot 2^x \, dx = \frac{dt}{(\ln 2)^3}$.Therefore,$\int \frac{dt}{(\ln 2)^3} = \frac{t}{(\ln 2)^3} + c = \frac{2^{2^{2^x}}}{(\ln 2)^3} + c$.

If $c$ is any arbitrary constant,then $\int {{2^{{2^{{2^x}}}}}{2^{{2^x}}}{2^x}dx} $ is equal to

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