Question

If $\int 2^{2^{x}} \cdot 2^{x} d x=A \cdot 2^{2^{x}}+c$, then $A=$
(A) $\frac{1}{\log 2}$
(B) $\log 2$
(C) $(\log 2)^{2}$
(D) $\frac{1}{(\log 2)^{2}}$

Answer 1

Ans: (D)
Hint $: 2^{x}=z \Rightarrow 2^{x} d x=\frac{d z}{\ln 2}$
$$\Rightarrow \frac{1}{\ln 2} \int 2^{z} d z=\frac{1}{(\ln 2)^{2}} 2^{2^{x}}+c$$

Answer 2

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Answer 3

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