Discussion
\(\int \frac{\sec^{2}(\sqrt{2x + 5})}{\sqrt{2x + 5}}dx\)=
Answer: D
The integral is \(\tan(\sqrt{2x + 5}) + C\).
Put \(u = \sqrt{2x + 5}\); then \(du = \frac{dx}{\sqrt{2x + 5}}\), exactly the rest of the integrand:
\[ \int\sec^{2}u\,du = \tan u + C = \tan\sqrt{2x + 5} + C \]
The factors 2 and \(\frac{1}{2}\) cancel inside \(du\), so the options with 2 or \(\frac{1}{2}\) in front are wrong, and those in \(\tan(2x + 5)\) lose the square root.
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