Phương pháp giải:

Sử dụng công thức: \(f\left( x \right) = \int {f'\left( x \right)dx} .\)

Lời giải chi tiết:

Ta có:

\(\begin{array}{l}f\left( x \right) = \int {f'\left( x \right)} dx = \int {\left( {2{{\cos }^2}x + 1} \right)dx}  = \int {\left( {\cos 2x + 1 + 1} \right)dx} \\\,\,\,\,\,\,\,\,\,\,\,\,\, = \int {\left( {\cos 2x + 2} \right)dx}  = \frac{1}{2}\sin 2x + 2x + C.\end{array}\)

Lại có: \(f\left( 0 \right) = 4 \Leftrightarrow \frac{1}{2}\sin 0 + 0 + C = 4 \Leftrightarrow C = 4.\)

\(\begin{array}{l} \Rightarrow f\left( x \right) = \frac{1}{2}\sin 2x + 2x + 4.\\ \Rightarrow \int\limits_0^{\frac{\pi }{4}} {f\left( x \right)dx}  = \int\limits_0^{\frac{\pi }{4}} {\left( {\frac{1}{2}\sin 2x + 2x + 4} \right)dx}  = \left. {\left( { - \frac{1}{4}\cos 2x + {x^2} + 4x} \right)} \right|_0^{\frac{\pi }{4}}\\\,\,\, = \left( {\frac{{{\pi ^2}}}{{16}} + \pi } \right) - \left( { - \frac{1}{4}} \right) = \frac{{{\pi ^2} + 16\pi  + 4}}{{16}}.\end{array}\)

Chọn  C.