Question:
\(\int \frac{\sec^2(\sqrt{2x+5})}{\sqrt{2x+5}} \, dx =\)
\(\int \frac{\sec^2(\sqrt{2x+5})}{\sqrt{2x+5}} \, dx =\)
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\(\frac{d}{dx}(\sqrt{2x+5}) = \frac{1}{\sqrt{2x+5}}\).
Updated On: Apr 27, 2026
- \(2 \tan(\sqrt{2x+5}) + C\)
- \(\frac{1}{2} \tan(\sqrt{2x+5}) + C\)
- \(\tan(2x+5) + C\)
- \(\tan(\sqrt{2x+5}) + C\)
- \(2 \tan(2x+5) + C\)
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The Correct Option is D
Solution and Explanation
Step 1: Concept:
• Use substitution: \[ u = \sqrt{2x+5} \]
• Then: \[ du = \frac{1}{\sqrt{2x+5}} \, dx \]
Step 2: Detailed Explanation:
• Substitute into the integral: \[ \int \frac{\sec^2(\sqrt{2x+5})}{\sqrt{2x+5}} \, dx \]
• This becomes: \[ \int \sec^2 u \, du \]
• Integrate: \[ \int \sec^2 u \, du = \tan u + C \]
• Substitute back: \[ = \tan(\sqrt{2x+5}) + C \]
Step 3: Final Answer:
• \[ \tan(\sqrt{2x+5}) + C \]
• Use substitution: \[ u = \sqrt{2x+5} \]
• Then: \[ du = \frac{1}{\sqrt{2x+5}} \, dx \]
Step 2: Detailed Explanation:
• Substitute into the integral: \[ \int \frac{\sec^2(\sqrt{2x+5})}{\sqrt{2x+5}} \, dx \]
• This becomes: \[ \int \sec^2 u \, du \]
• Integrate: \[ \int \sec^2 u \, du = \tan u + C \]
• Substitute back: \[ = \tan(\sqrt{2x+5}) + C \]
Step 3: Final Answer:
• \[ \tan(\sqrt{2x+5}) + C \]
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