A

To evaluate the integrals, we proceed as follows:
(i) Evaluation of \( \int (\sin x + \frac{1}{x} + 2 \frac{1}{x^2} + 3x^3) \, dx \):

• \( \int \sin x \, dx = -\cos x + C_1 \)
• \( \int \frac{1}{x} \, dx = \ln |x| + C_2 \)
• \( \int 2 \frac{1}{x^2} \, dx = -\frac{2}{x} + C_3 \)
• \( \int 3x^3 \, dx = \frac{3}{4} x^4 + C_4 \)

Combining all results, we have: \( -\cos x + \ln |x| - \frac{2}{x} + \frac{3}{4} x^4 + C \)

(ii) Evaluation of \( \int (3\cos x + e^x + 4x^2 + x + 5) \, dx \):

• \( \int 3\cos x \, dx = 3\sin x + C_5 \)
• \( \int e^x \, dx = e^x + C_6 \)
• \( \int 4x^2 \, dx = \frac{4}{3} x^3 + C_7 \)
• \( \int x \, dx = \frac{1}{2} x^2 + C_8 \)
• \( \int 5 \, dx = 5x + C_9 \)

Combining all results gives: \( 3\sin x + e^x + \frac{4}{3} x^3 + \frac{1}{2} x^2 + 5x + C \)

Therefore, the answer to the two integrals is:
1. \( -\cos x + \ln |x| - \frac{2}{x} + \frac{3}{4} x^4 + C \)
2. \( 3\sin x + e^x + \frac{4}{3} x^3 + \frac{1}{2} x^2 + 5x + C \) respectively.