Tính I = nguyên hàm 3 x^5 căn bậc hai của x^3 +1 dx

A. $I=\frac{1}{5}{\left(x^3+1\right)}^2\sqrt{x^3+1}-\frac{1}{3}\left(x^3+1\right)\sqrt{x^3+1}+C$

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

C. $I=\frac{2}{5}{\left(x^3+1\right)}^2\sqrt{x^3+1}+C$

D. $I=\frac{2}{5}{\left(x^3+1\right)}^2\sqrt{x^3+1}+\left(x^3+1\right)\sqrt{x^3+1}+C$