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$\int \frac{3^x}{\sqrt{1-9^x}} d x$ is equal to
Options:
Solution:
2832 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{\log 3} \sin ^{-1} 3^x+C$
We have,
$$
I=\int \frac{3^x}{\sqrt{1-9^x}} d x=\int \frac{3^x}{\sqrt{1-\left(3^x\right)^2}} d x
$$
Putting $3^x=t$
$$
\begin{array}{rlrl}
\Rightarrow \quad & 3^x \log 3 & d x=d t \\
\Rightarrow \quad & 3^x d x & =\frac{1}{\log 3} d t \\
\therefore \quad I & =\frac{1}{\log 3} \int \frac{1}{\sqrt{1-t^2}} d t \\
& & =\frac{1}{\log 3} \sin ^{-1} t+C \\
& = & \frac{1}{\log 3} \sin ^{-1}\left(3^x\right)+C
\end{array}
$$
Option (3) is correct.
$$
I=\int \frac{3^x}{\sqrt{1-9^x}} d x=\int \frac{3^x}{\sqrt{1-\left(3^x\right)^2}} d x
$$
Putting $3^x=t$
$$
\begin{array}{rlrl}
\Rightarrow \quad & 3^x \log 3 & d x=d t \\
\Rightarrow \quad & 3^x d x & =\frac{1}{\log 3} d t \\
\therefore \quad I & =\frac{1}{\log 3} \int \frac{1}{\sqrt{1-t^2}} d t \\
& & =\frac{1}{\log 3} \sin ^{-1} t+C \\
& = & \frac{1}{\log 3} \sin ^{-1}\left(3^x\right)+C
\end{array}
$$
Option (3) is correct.
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