$\int \frac{a^{x / 2}}{\sqrt{a^{-x}-a^x}} d x$ is equal to
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$\int \frac{a^{x / 2}}{\sqrt{a^{-x}-a^x}} d x$ is equal to

(a) $\frac{1}{\log a} \sin ^{-1}\left(a^x\right)+C$

(b) $\frac{1}{\log a} \tan ^{-1}\left(a^x\right)+C$

(c) $2 \sqrt{a^{-x}-a^x}+C$

(d) $\log \left(a^x-1\right)+C$

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Best answer

SOLUTION — Put $a^x=t \Rightarrow a^x \log a d x=d t$

$\begin{aligned}\therefore \quad I & =\frac{1}{\log a} \int \frac{d t}{\sqrt{1-t^2}} \\& =\frac{1}{\log a} \sin ^{-1} t+C \\& =\frac{1}{\log a} \sin ^{-1}\left(a^x\right)+C\end{aligned}$

So, The correct option of this question will be (A).

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