Question

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

• A. $\frac{1}{log3}log|3^x+\sqrt{9^x-1}|+c$
• B. $\frac{1}{log3}log|9^x+\sqrt{9^x-1}|+c$
• C. $\frac{1}{log9}log|3^x+\sqrt{9^x-1}|+c$
• D. $\frac{1}{log9}log|3^x-\sqrt{9^x-1}|+c$

Answer 1

The correct option is A $\frac{1}{log3}log|3^x+\sqrt{9^x-1}|+c$

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

$\Rightarrow$ Put $t=3^x$

$dt=3^{x}dx\cdot log3$

$\frac{dt}{log3}=3^{x}dx$

$=\frac{1}{log3}\int \frac{dt}{\sqrt{t^2-1}}$

$=\frac{1}{log3}log(t+\sqrt{t^2-1})+c$ ......... $\because \int \frac{1}{\sqrt{x^2-a^2}}dx=\log (x+\sqrt{x^2-a^2})+c$

$=\frac{1}{log3}log(3^x+\sqrt{9^x-1})+c$.