Question:medium

\( \int \frac{e^{2025x} + e^{-2025x}}{e^{2025x} + e^{-2025x}} dx = \)_____ + C

Show Hint

If numerator resembles derivative of denominator, use substitution immediately.
Updated On: Apr 2, 2026
  • \( e \log|e^x + e^{-x}| \)
  • \( \frac{1}{e} \log|e^x + e^{-x}| \)
  • \( \log|e^x + e^{-x}| \)
  • \( -\frac{1}{e} \log|e^x + e^{-x}| \)
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
Observe the integrand. The numerator and denominator are identical: $e^{2025x} + e^{-2025x}$.
Step 2: Formula Derivation:
Any expression divided by itself is equal to 1 (provided it is non-zero). $$\frac{e^{2025x} + e^{-2025x}}{e^{2025x} + e^{-2025x}} = 1$$ The integral becomes: $$\int 1 \, dx$$
Step 3: Explanation:
The integral of 1 with respect to $x$ is simply $x + C$. Note: The options provided seem to correspond to a different question (likely $\int \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} dx$). For the question as written, the answer is $x + C$.
Step 4: Final Answer:
The integral evaluates to $x + C$.
Was this answer helpful?
0