Question:
$\int \frac{\text{d}x}{2\text{e}^{2x}+3\text{e}^x+1} =$
$\int \frac{\text{d}x}{2\text{e}^{2x}+3\text{e}^x+1} =$
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Use substitution \(e^x = t\) to simplify exponential integrals.
Updated On: Apr 26, 2026
- $x + \log (\text{e}^x + 1) - 2 \log (2\text{e}^x + 1) + \text{c}$
- $x - \log (\text{e}^x + 1) + 4 \log (\text{e}^x + 1) + \text{c}$
- $x + \log (\text{e}^x + 1) - 4 \log (2\text{e}^x + 1) + \text{c}$
- $x - \log (\text{e}^x + 1) + 2 \log (2\text{e}^x + 1) + \text{c}$
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The Correct Option is A
Solution and Explanation
Concept:
Substitute: \[ t = e^x \] Step 1: Transform integral. \[ \int \frac{dx}{2e^{2x}+3e^x+1} = \int \frac{dt}{t(2t^2+3t+1)} \]
Step 2: Factor. \[ 2t^2+3t+1 = (2t+1)(t+1) \]
Step 3: Partial fractions. Solve and integrate.
Step 4: Conclusion. Answer matches option (A).
Substitute: \[ t = e^x \] Step 1: Transform integral. \[ \int \frac{dx}{2e^{2x}+3e^x+1} = \int \frac{dt}{t(2t^2+3t+1)} \]
Step 2: Factor. \[ 2t^2+3t+1 = (2t+1)(t+1) \]
Step 3: Partial fractions. Solve and integrate.
Step 4: Conclusion. Answer matches option (A).
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