Class CBSE Class 12 Mathematics Integrals Q #1703
COMPETENCY BASED
APPLY
1 Marks 2026 AISSCE(Board Exam) MCQ SINGLE
$\int\frac{dx}{2^{x}+2^{-x}}$ 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) $(\log 2)\tan^{-1}(2^{x})+C$
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Correct Answer: C

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Detailed Solution

Step 1: Simplify the integrand

The given integral is $I = \int \frac{dx}{2^x + 2^{-x}}$. We can rewrite $2^{-x}$ as $\frac{1}{2^x}$. Thus, the expression becomes: $$I = \int \frac{dx}{2^x + \frac{1}{2^x}} = \int \frac{2^x dx}{(2^x)^2 + 1}$$

Step 2: Substitution

Let $u = 2^x$. Then, the derivative is $du = 2^x \ln(2) dx$. This implies $2^x dx = \frac{du}{\ln(2)}$.

Step 3: Evaluate the integral

Substituting $u$ and $du$ into the integral: $$I = \int \frac{1}{u^2 + 1} \cdot \frac{du}{\ln(2)} = \frac{1}{\ln(2)} \int \frac{du}{u^2 + 1}$$ Using the standard integral formula $\int \frac{du}{u^2 + 1} = \tan^{-1}(u) + C$, we get: $$I = \frac{1}{\ln(2)} \tan^{-1}(u) + C$$

Step 4: Back-substitution

Replace $u$ with $2^x$: $$I = \frac{\tan^{-1}(2^x)}{\ln(2)} + C$$

Final Answer: $\frac{\tan^{-1}(2^{x})}{\log 2}+C$

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because the student must identify the appropriate substitution method to transform a non-standard integral into a standard form.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the execution of a specific sequence of mathematical steps (algebraic manipulation followed by integration by substitution).
Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. This tests the student's ability to handle exponential functions within the chapter of Integrals, moving beyond rote memorization of basic formulas.
||KEY:C

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