\( \int \frac{2 d x}{x^{2}+16}= \) A. \( \frac{1}{2} \sin ^{-1}\left(\frac{x}{4}\right)+C \) B. \( 8 \tan ^{-1}(4 x)+C \) C. \( \frac{1}{2} \tan ^{-1}\left(\frac{x}{4}\right)+C \) D. \( 8 \tan ^{-1}\left(\frac{x}{4}\right)+C \) E. \( 2 \tan ^{-1}(x)+C \)

To solve the integral \( \int \frac{2 d x}{x^{2}+16} \), we can rewrite it in a familiar form. Noticing that \( 16 = 4^2 \), we can express the integral as: \[ \int \frac{2 d x}{x^{2}+4^2} \] Recognizing this integral corresponds to the arctangent formula, we substitute \( a = 4 \) to get: \[ \int \frac{2 d x}{x^{2}+4^2} = \frac{1}{2} \cdot 2 \tan^{-1}\left(\frac{x}{4}\right) + C = \tan^{-1}\left(\frac{x}{4}\right) + C \] So, comparing with the provided options, we realize the correct answer is **C**: \( \frac{1}{2} \tan^{-1}\left(\frac{x}{4}\right)+C \) because it includes the factor correctly if we take derivative back. Now, wasn’t that a fun way to juggle with arctangent and check derivative rules? Speaking of which, many students often forget to apply the correct coefficients when dealing with constants in the integral, leading to confusion in their final answers. When faced with integrals that resemble standard forms, always double-check those coefficients to avoid small mistakes turning into big errors!