Question

$\int\frac{3 ^{x}}{\sqrt{1-9 ^{x}}}dx\quad$is equal to

• A. $\left(log \, 3\right)sin^{-1}\left(3^{x}\right)+C$
• B. $\frac{1}{3}sin^{-1}\left(3^{x}\right)+C$
• C. $\left(\frac{1}{log 3}\right)sin^{-1}\left(3^{x}\right)+C$
• D. $\frac{1}{9}sin^{-1}\left(3^{x}\right)+C$
• E. $sin^{-1}\left(3^{x}\right)+C$

Answer 1

Answer: (C)

Let $I=\int \frac{3^{n}}{\sqrt{1-9^{x}}} \,d x$
$\Rightarrow \, I=\int \frac{3^{x}}{\sqrt{1-\left(3^{\times}\right)^{2}}} d x$
put $ t=3^{x} $
$d t=3^{x} \log 3 \cdot d x$
Then, $I=\frac{1}{\log 3} \int \frac{d t}{\sqrt{1-t^{2}}}$
$\Rightarrow \, I=\frac{1}{\log 3} \cdot \sin ^{-1} t+C$
$\Rightarrow \, I=\left(\frac{1}{\log 3}\right) \cdot \sin ^{-1}\left(3^{x}\right)+C$