∫3x9x−1dx is equal to - Infinity Learn by Sri Chaitanya

A
$\frac{1}{\log 3} \log |3^{x} + \sqrt{9^{x} - 1}| + C$
B
$\frac{1}{\log 3} \log |9^{x} + \sqrt{9^{x} - 1}| + C$
C
$\frac{1}{\log 9} \log |3^{x} + \sqrt{9^{x} - 1}| + C$
D
$\frac{1}{\log 9} \log |3^{x} - \sqrt{9^{x} - 1}| + C$

Solution:

Let $I = \int \frac{3^{x}}{\sqrt{9^{x} - 1}} dx = \int \frac{3^{x} dx}{\sqrt{{(3^{x})}^{2} - 1}}$

put , $3^{x} = t \Rightarrow 3^{x} \log 3 dx = dt \Rightarrow 3^{x} dx = \frac{dt}{\log 3}$

$\begin{array}{l} ∴ I = \frac{1}{\log 3} \int \frac{dt}{\sqrt{t^{2} - 1}} \\ = \frac{1}{\log 3} \log [t + \sqrt{t^{2} - 1}] + C \\ = \frac{1}{\log 3} \log [3^{x} + \sqrt{9^{x} - 1}] + C \end{array}$

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

Solution:

Related content