Question

Evaluate the integral: $\int \frac{10^{x/2}}{\sqrt{10^{-x} - 10^x}} dx$

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

Hint:

Whenever you see an integral structured like $\frac{f'(x)}{\sqrt{1 - [f(x)]^2}}$, it will always integrate directly into an arc-sine structure: $\sin^{-1}(f(x))$. Since the derivative of an exponential function $10^x$ introduces a scaling factor of $\log 10$ in the numerator, it must appear as a reciprocal $\frac{1}{\log 10}$ in the front of your final integrated answer.

Answer 1

The Correct Option is C

Step 1: Clean up the denominator.
Write $10^{-x} = \dfrac{1}{10^x}$, so the radicand is $\dfrac{1}{10^x} - 10^x = \dfrac{1 - (10^x)^2}{10^x}$.
Step 2: Pull the power out of the root.
$\sqrt{\dfrac{1-(10^x)^2}{10^x}} = \dfrac{\sqrt{1-(10^x)^2}}{10^{x/2}}$, so the integrand becomes $\dfrac{10^{x/2}\cdot 10^{x/2}}{\sqrt{1-(10^x)^2}} = \dfrac{10^x}{\sqrt{1-(10^x)^2}}$.
Step 3: Choose a substitution.
Let $u = 10^x$. Then $du = 10^x\log 10\,dx$, so $10^x\,dx = \dfrac{du}{\log 10}$.
Step 4: Rewrite the integral in u.
$I = \dfrac{1}{\log 10}\displaystyle\int \dfrac{du}{\sqrt{1-u^2}}$.
Step 5: Use the standard formula.
$\int \dfrac{du}{\sqrt{1-u^2}} = \sin^{-1}u$, so $I = \dfrac{1}{\log 10}\sin^{-1}u + c$.
Step 6: Substitute back.
Replacing $u = 10^x$ gives $I = \dfrac{1}{\log 10}\sin^{-1}(10^x) + c$, which is option 3 and matches the key.
\[ \boxed{\dfrac{1}{\log 10}\sin^{-1}(10^x) + c} \]