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The value of $\int \frac{10^{x / 2}}{\sqrt{10^{-x}-10^{x}}} d x$ is
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Verified Answer
The correct answer is:
$\frac{1}{\log _{\varepsilon} 10} \sin ^{-1}\left(10^{x}\right)+c$
Let $\begin{aligned} I &=\int \frac{10^{x / 2}}{\sqrt{10^{-x}-10^{x}}} d x \\ &=\int \frac{10^{x / 2}}{\sqrt{\frac{1}{10^{x}}-10^{x}}} d x=\int \frac{10^{x / 2} \cdot 10^{x / 2}}{\sqrt{1-\left(10^{x}\right)^{2}}} d x \\ &=\int \frac{10^{x}}{\sqrt{1-\left(10^{x}\right)^{2}}} d x \end{aligned}$
Put $10^{x}=t \Rightarrow 10^{x} \log 10 d x=d t$
$$
\begin{aligned}
\mathrm{So}, I &=\frac{1}{\log 10} \int \frac{d t}{\sqrt{1-t^{2}}} \\
&=\frac{1}{\log 10} \sin ^{-1}(t)+c=\frac{1}{\log 10} \sin ^{-1}\left(10^{x}\right)+c
\end{aligned}
$$
Put $10^{x}=t \Rightarrow 10^{x} \log 10 d x=d t$
$$
\begin{aligned}
\mathrm{So}, I &=\frac{1}{\log 10} \int \frac{d t}{\sqrt{1-t^{2}}} \\
&=\frac{1}{\log 10} \sin ^{-1}(t)+c=\frac{1}{\log 10} \sin ^{-1}\left(10^{x}\right)+c
\end{aligned}
$$
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