Cho hàm số y = f(x) thỏa mãn $\text{f}\left({\text{x}}^3+3\text{x}+1\right)=3\text{x}+2 , \forall \text{x}\in ℝ$. Tích phân $\text{I}={\int }_1^5\text{x}{\text{f}}^{\text{'}}\left(\text{x}\right)\text{dx}$ có kết quả dạng $\frac{\text{a}}{\text{b}}$. Tính a + b

Đặt $\left\{\begin{array}{l}u=x\\ dv=f^{\text{'}}\left(x\right)dx\end{array}\Rightarrow \left\{\begin{array}{l}du=dx\\ v=f\left(x\right)\end{array}\right.\right.$

Khi đó $I={\left.xf\left(x\right)\right|}_1^5-{\int }_1^{5}f\left(x\right)dx$. Từ $f\left(x^3+3x+1\right)=3x+2\iff \left\{\begin{array}{l}f\left(5\right)=5(x=1)\\ f\left(1\right)=2(x=0)\end{array}\right.$

Suy ra $I=23-{\int }_1^{5}f\left(t\right)dt$, đặt $t=x^3+3x+1\Rightarrow \left\{\begin{array}{l}dt=\left(3x^2+3\right)dx\\ f\left(t\right)=3x+2\end{array}\right.$

Đổi cận, với $t=1\Rightarrow 1=x^3+3x+1\iff x=0$ và $t=5\Rightarrow 5=x^3+3x+1\iff x=1$