test thôi :)) nếu sai xin chỉ bảo$I=4\int\limits_{1}^{8}xdx-\frac{1}{3}\int\limits_{1}^{8}\frac{dx}{\sqrt[3]{x^2}}$$=2x^2|^8_1-\frac{1}{3}\int\limits_{1}^{8}\frac{dx}{\sqrt[3]{x^2}}=126-\frac{1}{3}\int\limits_{1}^{8}\frac{dx}{\sqrt[3]{x^2}}$Gọi $I_1$ là tích phân thứ $2$Đặt $\begin{cases}u= \sqrt[3]{x^2}\\ dv=dx \end{cases}\Rightarrow \begin{cases}du=\frac{2}{3}x^{(\frac{-1}{3})}dx \\ v=x \end{cases}$$\Rightarrow I_1=x^{5/3}|^8_1-\frac{2}{3}\int\limits_{1}^{8}x^{2/3}dx$$I_1=31-\frac{2}{3}.\frac{x^{5/3}}{5/3}|^8_1=31-\frac{2}{5}x^{5/3}|^8_1=31-\frac{62}{5}=\frac{93}{5}$$\Rightarrow I=126-I_1=126-\frac{93}{5}=\frac{537}{5}$
$I=2x^2|^8_1-\frac{1}{3}\int\limits_{1}^{8}x^{(-
2/3)}dx$$=51
0-\frac{1}{3}.\frac{x^{
1/3}}{
1/3}|^8_1
$$=
51
0-\
sqr
t[3]{x}|^8_1=5
09$