Cho hàm số $f \left( x \right)$ có đạo hàm liên tục trên đoạn $\left[ 1 ; 8 \left]\right.$ và thỏa mãn
$\int_{1}^{2} \left(\left[\right. f \left(\right. x^{3} \right) \left]\right)^{2} \text{d} x + 2 \int_{1}^{2} f \left(\right. x^{3} \right) \text{d} x - \dfrac{4}{3} \int_{1}^{8} f \left( x \right) \text{d} x = - \dfrac{247}{15}$.
Giả sử rằng $F \left( x \right)$ là một nguyên hàm của hàm số $f \left( x \right)$ trên $\left[\right. 1 ; 8 \left]\right.$. Tích phân $\int_{1}^{8} x \cdot F^{'} \left( x \right) \text{d} x$ bằng

Xét $I = \int_{1}^{2} \left(\left[ f \left(\right. x^{3} \right) \left]\right)^{2} \text{d} x + 2 \int_{1}^{2} f \left(\right. x^{3} \right) \text{d} x = \int_{1}^{2} \left(\left[ f \left(\right. x^{3} \right) + 1 \left]\right.\right)^{2} \text{d} x - \int_{1}^{2} \text{d} x = \int_{1}^{2} \left(\left[ f \left(\right. x^{3} \right) + 1 \left]\right.\right)^{2} \text{d} x - 1$.
Đặt $t = x^{3} \Rightarrow \text{d} t = 3 x^{2} \text{d} x \Leftrightarrow \text{d} x = \dfrac{\text{d} t}{3 \sqrt[3]{t^{2}}}$.
Với $x = 1 \Rightarrow t = 1$;
$x = 2 \Rightarrow t = 8$.
Ta có $I = \dfrac{1}{3} \int_{1}^{8} \dfrac{\left(\left[\right. f \left( t \right) + 1 \left]\right.\right)^{2}}{\sqrt[3]{t^{2}}} \text{d} t - 1 = \dfrac{1}{3} \int_{1}^{8} \left(\left[\right. \dfrac{f \left( x \right) + 1}{\sqrt[3]{x}} \left]\right.\right)^{2} \text{d} x - 1$.
Do đó
$\int_{1}^{2} \left(\left[ f \left(\right. x^{3} \right) \left]\right)^{2} \text{d} x + 2 \int_{1}^{2} f \left(\right. x^{3} \right) \text{d} x - \dfrac{4}{3} \int_{1}^{8} f \left( x \right) \text{d} x = - \dfrac{247}{15}$
$\Leftrightarrow \dfrac{1}{3} \int_{1}^{8} \left(\left[\right. \dfrac{f \left( x \right) + 1}{\sqrt[3]{x}} \left]\right.\right)^{2} \text{d} x - 1 - \dfrac{4}{3} \int_{1}^{8} f \left( x \right) \text{d} x = \dfrac{- 247}{15}$
$\Leftrightarrow \int_{1}^{8} \left(\left[\right. \dfrac{f \left( x \right) + 1}{\sqrt[3]{x}} \left]\right.\right)^{2} \text{d} x - 3 - 4 \int_{1}^{8} f \left( x \right) \text{d} x = \dfrac{- 247}{5}$
$\Leftrightarrow \int_{1}^{8} \left[\right. \left(\left(\right. \dfrac{f \left( x \right) + 1}{\sqrt[3]{x}} \left.\right)\right)^{2} - 4 f \left( x \right) \left]\right. \text{d} x = \dfrac{- 232}{5}$
$\Leftrightarrow \int_{1}^{8} \left[\right. \left(\left(\right. \dfrac{f \left( x \right) + 1}{\sqrt[3]{x}} \left.\right)\right)^{2} - 2 \cdot \dfrac{f \left( x \right) + 1}{\sqrt[3]{x}} \cdot 2 \sqrt[3]{x} + 4 \sqrt[3]{x^{2}} \left]\right. \text{d} x + \int_{1}^{8} \left[\right. - 4 \sqrt[3]{x^{2}} + 4 \left]\right. \text{d} x = \dfrac{- 232}{5}$
$\Leftrightarrow \int_{1}^{8} \left(\left[\right. \dfrac{f \left( x \right) + 1}{\sqrt[3]{x}} - 2 \sqrt[3]{x} \left]\right.\right)^{2} \text{d} x = 0$, do $\int_{1}^{8} \left[\right. - 4 \sqrt[3]{x^{2}} + 4 \left]\right. \text{d} x = \dfrac{- 232}{5}$
$\Rightarrow \dfrac{f \left( x \right) + 1}{\sqrt[3]{x}} - 2 \sqrt[3]{x} = 0 \Rightarrow f \left( x \right) = 2 \sqrt[3]{x^{2}} - 1 = F^{'} \left( x \right)$.
Suy ra $\int_{1}^{8} x \cdot F^{'} \left( x \right) \text{d} x = \int_{1}^{8} x \left[ 2 \sqrt[3]{x^{2}} - 1 \left]\right. \text{d} x$ $= 2 \int_{1}^{8} x^{\dfrac{5}{3}} \text{d} x - \int_{1}^{8} x \text{d} x = \dfrac{639}{4}$.