Question

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

• A. $\left(log3\right)si{n}^{-1}\left(3^x\right)+C$
• B. $\frac{1}{3}si{n}^{-1}\left(3^x\right)+C$
• C. $\left(\frac{1}{log3}\right)si{n}^{-1}\left(3^x\right)+C$
• D. $\frac{1}{9}si{n}^{-1}\left(3^x\right)+C$
• E. $si{n}^{-1}\left(3^x\right)+C$

Answer 1

Answer: (C)

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