Question

[Pham Trong Bach]

Cho $\mathrm{F}\left(\mathrm{x}\right)$ là một nguyên hàm của hàm số $\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}\sqrt{\mathrm{x}+1}$. Tìm $\mathrm{F}\left(\mathrm{x}\right)$ biết  $\mathrm{F}\left(0\right)=0$

A. $\mathrm{F}\left(\mathrm{x}\right)=\frac{2}{5}\sqrt{{\left(\mathrm{x}+1\right)}^5}-\frac{2}{3}\sqrt{{\left(\mathrm{x}+1\right)}^3}+\frac{4}{15}$

B.  $\mathrm{F}\left(\mathrm{x}\right)=\frac{1}{5}\sqrt{{\left(\mathrm{x}+1\right)}^5}-\frac{1}{3}\sqrt{{\left(\mathrm{x}+1\right)}^3}+\frac{2}{15}$

C.  $\mathrm{F}\left(\mathrm{x}\right)=\frac{1}{5}\sqrt{{\left(\mathrm{x}+1\right)}^3}-\frac{7}{15}\sqrt{\mathrm{x}+1}+\frac{4}{15}$

D.  $\mathrm{F}\left(\mathrm{x}\right)=\frac{2}{5}\sqrt{{\left(\mathrm{x}+1\right)}^3}-\frac{2}{3}\sqrt{\mathrm{x}+1}+\frac{4}{15}$

Answer 1

[Cao Minh Tâm]