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If $\int 2^{2^{x}} \cdot 2^{x} d x=A \cdot 2^{2^{x}}+C,$ then $A$ is equal to
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1185 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{(\log 2)^{2}}$
$$
\begin{aligned}
&\text { (d) Let } I=\int 2^{2^{x}} \cdot 2^{x} d x\\
&\begin{array}{l}
\text { Put } \quad t=2^{x} \\
\Rightarrow \quad d t=2^{x} \log 2 d x \\
\therefore \quad I=\int \frac{2^{t}}{\log 2} d t \\
=\frac{2^{t}}{(\log 2)^{2}}+C=\frac{2^{2^{x}}}{(\log 2)^{2}}+C \\
\therefore \quad A=\frac{1}{(\log 2)^{2}}
\end{array}
\end{aligned}
$$
\begin{aligned}
&\text { (d) Let } I=\int 2^{2^{x}} \cdot 2^{x} d x\\
&\begin{array}{l}
\text { Put } \quad t=2^{x} \\
\Rightarrow \quad d t=2^{x} \log 2 d x \\
\therefore \quad I=\int \frac{2^{t}}{\log 2} d t \\
=\frac{2^{t}}{(\log 2)^{2}}+C=\frac{2^{2^{x}}}{(\log 2)^{2}}+C \\
\therefore \quad A=\frac{1}{(\log 2)^{2}}
\end{array}
\end{aligned}
$$
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