$10$ is a number. It formed three exponential terms $10^{x/2}$, $10^{-x}$ and $10^x$ as a common base of them. The three exponential terms formed a fractional function which also contains a square root. The integral of the function is solved to find its integration.

$\displaystyle \int$ $\dfrac{10^{\frac{x}{2}}}{\sqrt{10^{-x} -10^x}} \,\,dx$

The exponent of the number $10$ in the numerator is $\dfrac{x}{2}$ and it can be expressed as the product of $x$ and $\dfrac{1}{2}$.

$=$ $\displaystyle \int$ $\dfrac{10^{(x \times \frac{1}{2})}}{\sqrt{10^{-x} -10^x}} \,\,dx$

According to the power rule of exponents, the product of two exponents of a number can be expressed as the power of an exponential term.

$=$ $\displaystyle \int$ $\dfrac{{(10^x)}^{\frac{1}{2}}}{\sqrt{10^{-x} -10^x}} \,\,dx$

The power of $\dfrac{1}{2}$ of a number represents the square root of that number.

$=$ $\displaystyle \int$ $\dfrac{\sqrt{10^x}}{\sqrt{10^{-x} -10^x}} \,\,dx$

Now, simplify the denominator to transform the term $10^{-x}$ inside the square root in the denominator in terms of $10^x$.

$=$ $\displaystyle \int$ $\dfrac{\sqrt{10^x}}{\sqrt{\dfrac{1}{10^x} -10^x}} \,\,dx$

$=$ $\displaystyle \int$ $\dfrac{\sqrt{10^x}}{\sqrt{\dfrac{1-10^x \times 10^x}{10^x}}} \,\,dx$

$=$ $\displaystyle \int$ $\dfrac{\sqrt{10^x}}{\dfrac{\sqrt{1-{(10^x)}^2}}{\sqrt{10^x}}} \,\,dx$

$=$ $\displaystyle \int$ $\dfrac{\sqrt{10^x} \times \sqrt{10^x}}{\sqrt{1-{(10^x)}^2}} \,\,dx$

$=$ $\displaystyle \int$ $\dfrac{{(\sqrt{10^x})}^2}{\sqrt{1-{(10^x)}^2}} \,\,dx$

$=$ $\displaystyle \int$ $\dfrac{10^x}{\sqrt{1-{(10^x)}^2}} \,\,dx$

The integral of the function is in terms of $10^x$ completely. Take $m = 10^x$ and differentiate the equation by using differentiation of the $a^x$ formula.

$\implies dm = 10^x \log_{e} 10 \,\,dx$

$\implies \dfrac{dm}{\log_{e} 10} = 10^x \,\,dx$

$\implies 10^x \,\,dx = \dfrac{dm}{\log_{e} 10}$

Now, change the integral function in terms of $x$ as the integral function in terms of $m$ by the relation between them.

$\implies$ $\displaystyle \int$ $\dfrac{10^x}{\sqrt{1-{(10^x)}^2}} \,\,dx$ $=$ $\displaystyle \int$ $\dfrac{1}{\sqrt{1-m^2}} \dfrac{dm}{\log_{e} 10}$

$=$ $\displaystyle \int$ $\dfrac{1}{\sqrt{1-m^2}} \times \dfrac{1}{\log_{e} 10} \,\,dm$

The natural logarithm of $10$ is a constant. Hence, it can be taken out from the integral function.

$=$ $\dfrac{1}{\log_{e} 10}$ $\displaystyle \int$ $\dfrac{1}{\sqrt{1-m^2}} \,\,dm$

According to integral formula of $\dfrac{1}{\sqrt{1-x^2}}$ is equal to $sin^{-1} x$.

$=$ $\dfrac{1}{\log_{e} 10} [\sin^{-1} m + c] $

Now, transform the solution in terms of $x$ from $m$.

$=$ $\dfrac{1}{\log_{e} 10} [\sin^{-1} (10^x) + c] $

$\dfrac{1}{\log_{e} 10} \sin^{-1} (10^x) + \dfrac{1}{\log_{e} 10} \times c$

The product of $\dfrac{1}{\log_{e} 10}$ and $c$ is also constant value. Hence, it is simplify expressed as $c$.

$\displaystyle \int$ $\dfrac{10^{\frac{x}{2}}}{\sqrt{10^{-x} -10^x}} \,\,dx$ $=$ $\dfrac{1}{\log_{e} 10} \sin^{-1} (10^x) + c$

It is the required solution for this integration problem but it can also be writen as follows according to logarithms.

$\displaystyle \int$ $\dfrac{10^{\frac{x}{2}}}{\sqrt{10^{-x} -10^x}} \,\,dx$ $=$ $\dfrac{1}{\ln 10} \sin^{-1} (10^x) + c$

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