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$\int \frac{3^x}{\sqrt{9^x-1}} d x=$
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Verified Answer
The correct answer is:
$\frac{1}{\log 3} \log \left|3^x+\sqrt{9^x-1}\right|+c$
We have, $\int \frac{3^x}{\sqrt{9^x-1}} d x$
Put $3^x=t \Rightarrow 3^x \log 3 d x=d t$
$$
\begin{aligned}
& \therefore \frac{1}{\log 3} \int \frac{d t}{\sqrt{t^2-1}}=\frac{1}{\log 3} \log \left|t+\sqrt{t^2-1}\right|+c \\
& \text { Put } t=3^x=\frac{1}{\log 3} \log \left|3^x+\sqrt{9^x-1}\right|+c
\end{aligned}
$$
Put $3^x=t \Rightarrow 3^x \log 3 d x=d t$
$$
\begin{aligned}
& \therefore \frac{1}{\log 3} \int \frac{d t}{\sqrt{t^2-1}}=\frac{1}{\log 3} \log \left|t+\sqrt{t^2-1}\right|+c \\
& \text { Put } t=3^x=\frac{1}{\log 3} \log \left|3^x+\sqrt{9^x-1}\right|+c
\end{aligned}
$$
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