What is \int\frac{dx}{10^{x}+10^{-x}} equal to?
- A. \tan^{-1}(10^{x})+c
- B. (\ln 10)\tan^{-1}(10^{x})+c
- C. \frac{1}{\ln 10}\tan^{-1}(10^{x})+c ✓
- D. \ln(10^{x}+10^{-x})+c
Correct Answer: C. \frac{1}{\ln 10}\tan^{-1}(10^{x})+c
Explanation
Multiply numerator and denominator by 10^x to get \int \frac{10^x}{10^{2x}+1} \,dx. Let 10^x = t, so 10^x \ln 10 \,dx = dt, or dx = \frac{dt}{t \ln 10}. Substituting gives \frac{1}{\ln 10} \int \frac{1}{t^2+1} \,dt = \frac{1}{\ln 10} \tan^{-1}(t) + c. Replacing t yields \frac{1}{\ln 10}\tan^{-1}(10^x) + c.
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