We are given the integral:
\[
\int \frac{\sec^2(\sqrt{2x+5})}{\sqrt{2x+5}} \, dx
\]
To solve this, let's substitute \( u = \sqrt{2x + 5} \). This implies:
\[
du = \frac{1}{2\sqrt{2x + 5}} \cdot 2 \, dx = \frac{1}{\sqrt{2x+5}} \, dx
\]
Thus, the integral becomes:
\[
\int \sec^2(u) \, du = \tan(u) + C
\]
Now, substituting \( u = \sqrt{2x+5} \), we get:
\[
\tan(\sqrt{2x+5}) + C
\]
The correct answer is \( \frac{1{2} \log |2x + 5|} \).