If int 2^{2^{x}} cdot 2^{x} , dx = A cdot 2^{ | vedclass

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(D) Let $I = \int 2^{2^{x}} \cdot 2^{x} \, dx$.Substitute $t = 2^{x}$.Then, $dt = 2^{x} \ln 2 \, dx$, which implies $2^{x} \, dx = \frac{dt}{\ln 2}$.S

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$\frac{1}{(\log 2)^{2}}$

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(D) Let $I = \int 2^{2^{x}} \cdot 2^{x} \, dx$.Substitute $t = 2^{x}$.Then, $dt = 2^{x} \ln 2 \, dx$, which implies $2^{x} \, dx = \frac{dt}{\ln 2}$.Substituting these into the integral, we get:$I = \int 2^{t} \cdot \frac{dt}{\ln 2} = \frac{1}{\ln 2} \int 2^{t} \, dt$.Using the formula $\int a^{t} \, dt = \frac{a^{t}}{\ln a} + C$, we have:$I = \frac{1}{\ln 2} \cdot \frac{2^{t}}{\ln 2} + C = \frac{2^{t}}{(\ln 2)^{2}} + C$.Substituting back $t = 2^{x}$, we get:$I = \frac{2^{2^{x}}}{(\log 2)^{2}} + C$.Comparing this with $A \cdot 2^{2^{x}} + C$, we find $A = \frac{1}{(\log 2)^{2}}$.

If $\int 2^{2^{x}} \cdot 2^{x} \, dx = A \cdot 2^{2^{x}} + C,$ then $A$ is equal to

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