Tìm nguyên hàm \(F\left( x \right)\) của hàm số \(f\left( x \right) = {\cos ^2}\frac{x}{2}\)

A. \(\frac{4}{{15}}x\sqrt[{15}]{{{x^7}}} + C\).

B. \(\frac{8}{{15}}x\sqrt[{15}]{{{x^7}}} + C\).

C. \(\frac{8}{{15}}x\sqrt[{15}]{x} + C\).

D. \(\frac{4}{{15}}x\sqrt[{15}]{x} + C\).

A. \(\int {f\left( x \right)} dx = \frac{3}{2}{x^{\frac{1}{2}}} + C\).

B. \(\int {f\left( x \right)} dx = \int {\sqrt {{x^3}} } dx\).

C. \(\int {f\left( x \right)} dx = \frac{2}{5}{x^{\frac{5}{2}}} + C\).

D. \(\int {f\left( x \right)} dx = \frac{2}{3}{x^{\frac{1}{2}}} + C\).

A. \[{x^2} + C\].

B. \[{x^2} + 6x + C\].

C. \[2{x^2} + C\].

D. \[2{x^2} + 6x + C\].

A. \[\frac{1}{{12}}{x^4} - \frac{2}{3}{x^3} + \frac{{{x^2}}}{2} + C\].

B. \(\frac{1}{9}{x^4} - \frac{2}{3}{x^3} + \frac{{{x^2}}}{2} - 2024x + C\).

C. \(\frac{1}{{12}}{x^4} - \frac{2}{3}{x^3} + \frac{{{x^2}}}{2} - 2024x + C\).

D. \(\frac{1}{9}{x^4} + \frac{2}{3}{x^3} - \frac{{{x^2}}}{2} - 2024x + C\).

A. \[x\sqrt[5]{x} - 2x\sqrt[{17}]{{{x^5}}} + \sqrt[4]{{{x^3}}} + C\].

B. \[\frac{4}{5}x\sqrt[5]{x} - \frac{{24}}{{17}}x\sqrt[{17}]{{{x^5}}} + \frac{4}{3}\sqrt[4]{{{x^3}}} + C\].

C. \[x\sqrt[5]{x} - \frac{{24}}{{17}}x\sqrt[{17}]{{{x^5}}} + \sqrt[4]{{{x^3}}} + C\].

D. \[\frac{4}{5}x\sqrt[5]{x} - 2x\sqrt[{17}]{{{x^5}}} + \frac{4}{3}\sqrt[4]{{{x^3}}} + C\].

A. \(\frac{{{x^{2023}}}}{{2023}} + 1\).

B. \(\frac{{{x^{2023}}}}{{2023}}\).

C. \(y = 2022{x^{2021}}\).

D. \(\frac{{{x^{2023}}}}{{2023}} - 1\).

A. \(f\left( x \right) = 6{x^2} - 2\)

B. \(f\left( x \right) = \frac{1}{2}{x^4} - {x^2} + x\)

C. \(f\left( x \right) = \frac{1}{2}{x^4} - {x^2} + x + C\).

D. \(f\left( x \right) = 6{x^2} - 2 + C\).