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NDA 2025
Exam
NDA
Year
2025
Subject
Mathematics
Chapter
Indefinite Integration

NDA 2025 · Mathematics · Indefinite Integration

What is ∫ dx 10^x + 10^ -x equal to?

What is ∫dx10x+10−x\int \dfrac{dx}{10^x + 10^{-x}} equal to?

Options

  1. Atan⁡−1(10x)+c\tan^{-1}(10^x) + c
  2. B(ln⁡10)tan⁡−1(10x)+c(\ln 10)\tan^{-1}(10^x) + c
  3. C1ln⁡10tan⁡−1(10x)+c\dfrac{1}{\ln 10}\tan^{-1}(10^x) + c
  4. Dln⁡(10x+10−x)+c\ln(10^x + 10^{-x}) + c

Correct answer

C. 1ln⁡10tan⁡−1(10x)+c\dfrac{1}{\ln 10}\tan^{-1}(10^x) + c

Step-by-step solution

Let I=∫dx10x+10−xI = \int \dfrac{dx}{10^x + 10^{-x}}

I=∫10x102x+1dxI = \int \dfrac{10^x}{10^{2x} + 1} dx

Let 10x=t10^x = t

10xln⁡10 dx=dt⇒10x dx=dtln⁡1010^x \ln 10 \, dx = dt \Rightarrow 10^x \, dx = \dfrac{dt}{\ln 10}

I=1ln⁡10∫dtt2+1I = \dfrac{1}{\ln 10} \int \dfrac{dt}{t^2 + 1}

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

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

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