Question:
Evaluate the integral: $\int \frac{10^{x/2}}{\sqrt{10^{-x} - 10^x}} dx$
Evaluate the integral: $\int \frac{10^{x/2}}{\sqrt{10^{-x} - 10^x}} dx$
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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.
Updated On: Jun 12, 2026
- $2\sqrt{10^{-x} + 10^x} + c$
- $2\sqrt{2}\sqrt{10^{-x} + 10^x} + c$
- $\frac{1}{\log 10} \sin^{-1}\left(10^x\right) + c$
- $\frac{1}{\log 10} \cos^{-1}\left(10^x\right) + c$
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Question:
The problem requires finding the indefinite integral of an exponential fraction containing a radical expression in the denominator.
Step 2: Key Formula or Approach:
We can simplify the integrand by rewriting the negative exponent in the denominator as a fraction: $10^{-x} = \frac{1}{10^x}$. This allows us to use substitution to transform the expression into a standard inverse sine integral format: $$\int \frac{1}{\sqrt{1-u^2}} du = \sin^{-1}(u) + c$$
Step 3: Detailed Explanation:
Let the integral be $I$: $$I = \int \frac{10^{x/2}}{\sqrt{\frac{1}{10^x} - 10^x}} dx = \int \frac{10^{x/2}}{\sqrt{\frac{1 - (10^x)^2}{10^x}}} dx$$ Simplify the radical denominator fraction by pulling $10^x$ out of the square root as $10^{x/2}$: $$I = \int \frac{10^{x/2} \cdot 10^{x/2}}{\sqrt{1 - (10^x)^2}} dx = \int \frac{10^x}{\sqrt{1 - (10^x)^2}} dx$$ Now, apply integration by substitution. Let: $$u = 10^x \implies du = 10^x \log 10 \, dx \implies 10^x dx = \frac{1}{\log 10} du$$ Substitute these terms back into the integral: $$I = \frac{1}{\log 10} \int \frac{1}{\sqrt{1 - u^2}} du$$ Integrate using the standard inverse trigonometric formula: $$I = \frac{1}{\log 10} \sin^{-1}(u) + c$$ Substitute the original variable value $u = 10^x$ back to get the final solution: $$I = \frac{1}{\log 10} \sin^{-1}\left(10^x\right) + c$$
Step 4: Final Answer:
The solution to the indefinite integral matches option (C).
The problem requires finding the indefinite integral of an exponential fraction containing a radical expression in the denominator.
Step 2: Key Formula or Approach:
We can simplify the integrand by rewriting the negative exponent in the denominator as a fraction: $10^{-x} = \frac{1}{10^x}$. This allows us to use substitution to transform the expression into a standard inverse sine integral format: $$\int \frac{1}{\sqrt{1-u^2}} du = \sin^{-1}(u) + c$$
Step 3: Detailed Explanation:
Let the integral be $I$: $$I = \int \frac{10^{x/2}}{\sqrt{\frac{1}{10^x} - 10^x}} dx = \int \frac{10^{x/2}}{\sqrt{\frac{1 - (10^x)^2}{10^x}}} dx$$ Simplify the radical denominator fraction by pulling $10^x$ out of the square root as $10^{x/2}$: $$I = \int \frac{10^{x/2} \cdot 10^{x/2}}{\sqrt{1 - (10^x)^2}} dx = \int \frac{10^x}{\sqrt{1 - (10^x)^2}} dx$$ Now, apply integration by substitution. Let: $$u = 10^x \implies du = 10^x \log 10 \, dx \implies 10^x dx = \frac{1}{\log 10} du$$ Substitute these terms back into the integral: $$I = \frac{1}{\log 10} \int \frac{1}{\sqrt{1 - u^2}} du$$ Integrate using the standard inverse trigonometric formula: $$I = \frac{1}{\log 10} \sin^{-1}(u) + c$$ Substitute the original variable value $u = 10^x$ back to get the final solution: $$I = \frac{1}{\log 10} \sin^{-1}\left(10^x\right) + c$$
Step 4: Final Answer:
The solution to the indefinite integral matches option (C).
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