Question

The value of $\int \frac{10^{x/2}}{\sqrt{10^{-x}-10^x}}dx$ is

• A. $\frac{1}{{\log }_{e}10}{\sin }^{-1}\left(10^x\right)+c$
• B. $2\sqrt{10^{-x}+10^x}+c$
• C. $\frac{1}{{\log }_{e}10}{\sinh }^{-1}\left(10x\right)+c$
• D. $\frac{-1}{{\log }_{e}10}{\sinh }^{-1}\left(10x\right)+c$

Answer 1

Answer: (A)

Let $I=\int \frac{10^{\frac{x}{2}}}{\sqrt{10^{-x}-10^x}}dx=\int \frac{10^{\frac{x}{2}}}{\sqrt{\frac{1-10^{2x}}{10^x}}}dx$
$=\int \frac{10^{\frac{x}{2}}.10^{\frac{x}{2}}}{\sqrt{1-{\left(10^x\right)}^2}}dx=\int \frac{10^x}{\sqrt{1-{\left(10^x\right)}^2}}dx$
Put $10^x=t\Rightarrow 10^x\log 10dx=dt$
$\therefore I=\frac{1}{\log 10}\int \frac{dt}{\sqrt{1-t^2}}=\frac{1}{\log 10}\left({\sin }^{-1}t\right)+c$
$=\frac{1}{\log 10}{\sin }^{-1}\left(10^x\right)+c$