(Đề thi TNPT.2019 câu 32. Mã 101)

Lời giải   

\( f(x) = \int f'(x) \, dx = \int (2\cos^2 x + 1) \, dx = \int (2 + \cos 2x) \, dx \)

          \( = 2x + \frac{1}{2} \sin 2x + C \)

\( f(0) = 4 \implies C = 4 \implies f(x) = 2x + \frac{1}{2} \sin 2x + 4 \) 

\( \int_0^{\frac{\pi}{4}} f(x) \, dx = \int_0^{\frac{\pi}{4}} \left( 2x + \frac{1}{2} \sin 2x + 4 \right) \, dx =  x^2 - \frac{1}{4}\cos 2x  + 4x \Big|_0^{\frac{\pi}{4}} \)  

                       \( = \frac{\pi^2}{16} + \pi + \frac{1}{4} = \frac{\pi^2 + 16\pi + 4}{16} \)

\(\Rightarrow\) Vậy chọn đáp án \(\boxed{\text{C}} \)