Q103 / 200 · AP EAMCET 2002 · Mathematics

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

  • Correct answer
    A
    $\frac{1}{\log 3} \log \left|3^x+\sqrt{9^x-1}\right|+c$
  • B
    $\frac{1}{\log 3} \log \left|3^x-\sqrt{9^x-1}\right|+c$
  • C
    $\frac{1}{\log 9} \log \left|3^x+\sqrt{9^x-1}\right|+c$
  • D
    $\frac{1}{\log 3} \log \left|9^x+\sqrt{9^x-1}\right|+c$

Solution

Let $I=\int \frac{3^x d x}{\sqrt{9^x-1}}=\int \frac{3^x}{\sqrt{3^{2 x}-1}} d x$ Let $3^x=z \Rightarrow 3^x \log 3 d x=d z$ $$ I=\frac{1}{\log 3} \int \frac{z}{\sqrt{z^2-1}} $$ $\begin{aligned} & =\frac{1}{\log 3} \log \left\{z+\sqrt{z^2-1}\right\}+c \\ & =\frac{1}{\log 3} \log \left[3^x+\sqrt{9^x-1}\right]+c\end{aligned}$

Hence, the correct answer is option: 'A'.

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