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$\int \frac{4^x}{\sqrt{1-16^x}} d x$ is equal to
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Verified Answer
The correct answer is:
$\frac{1}{\log 4} \sin ^{-1} 4^x+C$
$\begin{aligned}
I & =\int \frac{4^x}{\sqrt{1-16^x}} d x \\
& =\int \frac{4^x}{\sqrt{1-\left(4^x\right)^2}} d x
\end{aligned}$
On putting, $4^x=t$, we get $4^x \log 4 d x=d t$
$\begin{aligned}
\Rightarrow \quad 4^x d x & =\frac{d t}{\log 4} \\
\therefore \quad I & =\frac{1}{\log 4} \int \frac{d t}{\sqrt{1-t^2}} \\
& =\frac{1}{\log 4} \sin ^{-1} t+C \\
& =\frac{1}{\log 4} \sin ^{-1} 4^x+C
\end{aligned}$
I & =\int \frac{4^x}{\sqrt{1-16^x}} d x \\
& =\int \frac{4^x}{\sqrt{1-\left(4^x\right)^2}} d x
\end{aligned}$
On putting, $4^x=t$, we get $4^x \log 4 d x=d t$
$\begin{aligned}
\Rightarrow \quad 4^x d x & =\frac{d t}{\log 4} \\
\therefore \quad I & =\frac{1}{\log 4} \int \frac{d t}{\sqrt{1-t^2}} \\
& =\frac{1}{\log 4} \sin ^{-1} t+C \\
& =\frac{1}{\log 4} \sin ^{-1} 4^x+C
\end{aligned}$
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