Question

\(\int \frac{dx}{2^x + 2^{-x}}\) is equal to :

• A. \(\tan^{-1} (2^x) + C\)
• B. \(\tan^{-1} (2^{-x}) + C\)
• C. \(\frac{\tan^{-1} (2^x)}{\log 2} + C\)
• D. \((\log 2) \tan^{-1} (2^x) + C\)

Hint:

Whenever an integrand contains \(a^x + a^{-x}\) in the denominator, multiply top and bottom by \(a^x\) to create the derivative of \(a^x\) in the numerator. Remember the scaling factor of \(\frac{1}{\log a}\) which arises from the differentiation of exponential terms with base other than \(e\).

Answer 1

The Correct Option is C

Concept:
• Rewrite negative exponents using fractions: \(2^{-x} = \frac{1}{2^x}\).
• Use substitution method for integration where \(u = a^x\) and \(du = a^x \ln a \, dx\).
• The standard integral formula: \(\int \frac{1}{u^2 + 1} \, du = \tan^{-1}(u) + C\).

Step 1: Simplify the integrand algebraically
Let the integral be \(I = \int \frac{dx}{2^x + 2^{-x}}\). Write \(2^{-x}\) as \(\frac{1}{2^x}\): \[ I = \int \frac{dx}{2^x + \frac{1}{2^x}} \] Multiply numerator and denominator by \(2^x\): \[ I = \int \frac{2^x}{(2^x)^2 + 1} \, dx \]

Step 2: Apply the method of substitution
Let \(u = 2^x\). Differentiating both sides with respect to \(x\): \[ \frac{du}{dx} = 2^x \log 2 \implies du = 2^x \log 2 \, dx \] Rearranging for \(2^x dx\): \[ 2^x dx = \frac{du}{\log 2} \]

Step 3: Integrate with respect to \(u\)
Substitute \(u\) and \(du\) into the integral: \[ I = \int \frac{\frac{du}{\log 2}}{u^2 + 1} = \frac{1}{\log 2} \int \frac{du}{u^2 + 1} \] Using the standard inverse tangent integration formula: \[ I = \frac{1}{\log 2} \tan^{-1}(u) + C \]

Step 4: Substitute back the original variable
Replace \(u\) with \(2^x\): \[ I = \frac{\tan^{-1}(2^x)}{\log 2} + C \]