What is ∫{dx}{10^{x}+10^{-x}} equal to?

Correct Answer is : 1ln10tan−1(10x)+c1/ln10tan^{-1}(10^{x})+cln101tan−1(10x)+c

Correct Answer is : 1ln10tan−1(10x)+c1/ln10tan^{-1}(10^{x})+cln101tan−1(10x)+c

Test Index

UPSC NDA II 2025 Mathematics Paper

Question : 95 of 120

Marks: +1, -0

What is

∫\frac{dx}{10^{x}+10^{-x}}

equal to?

Solution:

Concept:

Rewrite the integrand to form a standard integral of the form

\int \frac{dt}{1+t^2}

after a suitable substitution. Use the relationship between

\log_{10}e

and

\ln 10

Explanation:

First, multiply numerator and denominator by

10^{x}

\int \frac{10^{x}}{10^{2x}+1}dx

Substitute

t = 10^{x}

. Then

dt = 10^{x} \ln 10 \, dx

, so

dx = \frac{dt}{t \ln 10}

The integral becomes

\int \frac{t}{t^2+1} \cdot \frac{dt}{t \ln 10} = \frac{1}{\ln 10} \int \frac{dt}{1+t^2}

Integrate:

\frac{1}{\ln 10} \tan^{-1} t + c = \frac{1}{\ln 10} \tan^{-1}(10^{x}) + c

Note that

\log_{10}e = \frac{1}{\ln 10}

, confirming the answer.

Answer:

\frac{1}{\ln 10} \tan^{-1}(10^{x}) + c

corresponds to Option C.

Go to Question:

Prev Question

Next Question