$10$ 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.

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

Latest Math Topics

Latest Math Problems

A best free mathematics education website for students, teachers and researchers.

Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.

Learn how to solve the maths problems in different methods with understandable steps.

Copyright © 2012 - 2021 Math Doubts, All Rights Reserved