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$\int \frac{2^{x}}{\sqrt{1-4^{x}}} d x$ is equal to
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Verified Answer
The correct answer is:
$\frac{1}{\log 2} \sin ^{-1} 2^{x}+C$
Let $I=\int \frac{2^{x}}{\sqrt{1-4^{x}}} d x=\int \frac{2^{x}}{\sqrt{1-\left(2^{x}\right)^{2}}} d x$
Putting $2^{x}=t \Rightarrow 2^{x} \cdot \log 2 d x=d t$
$$
\begin{aligned}
\Rightarrow \quad 2^{x} d x &=\frac{d t}{\log 2} \\
\therefore \quad I &=\frac{1}{\log 2} \int \frac{d t}{\sqrt{1-t^{2}}}=\frac{1}{\log 2} \sin ^{-1}(t)+C \\
&=\frac{1}{\log 2} \sin ^{-1}\left(2^{x}\right)+C
\end{aligned}
$$
Putting $2^{x}=t \Rightarrow 2^{x} \cdot \log 2 d x=d t$
$$
\begin{aligned}
\Rightarrow \quad 2^{x} d x &=\frac{d t}{\log 2} \\
\therefore \quad I &=\frac{1}{\log 2} \int \frac{d t}{\sqrt{1-t^{2}}}=\frac{1}{\log 2} \sin ^{-1}(t)+C \\
&=\frac{1}{\log 2} \sin ^{-1}\left(2^{x}\right)+C
\end{aligned}
$$
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