Q245 / 270 · NDA 2025 (Phase 2) · Mathematics

What is $\int \dfrac{dx}{10^x + 10^{-x}}$ equal to?

  • A
    $\tan^{-1}(10^x) + c$
  • B
    $(\ln 10)\tan^{-1}(10^x) + c$
  • Correct answer
    C
    $\dfrac{1}{\ln 10}\tan^{-1}(10^x) + c$
  • D
    $\ln(10^x + 10^{-x}) + c$

Solution

Let $I = \int \dfrac{dx}{10^x + 10^{-x}}$

$I = \int \dfrac{10^x}{10^{2x} + 1} dx$

Let $10^x = t$

$10^x \ln 10 \, dx = dt \Rightarrow 10^x \, dx = \dfrac{dt}{\ln 10}$

$I = \dfrac{1}{\ln 10} \int \dfrac{dt}{t^2 + 1}$

$I = \dfrac{1}{\ln 10} \tan^{-1}(t) + c$

$I = \dfrac{1}{\ln 10} \tan^{-1}(10^x) + c$

Answer: $\dfrac{1}{\ln 10}\tan^{-1}(10^x) + c$

Hence, the correct answer is option: 'C'.

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