Cho hàm số ${f(x)=x^3+a x^2+b x+c}$ với ${a, b, c}$ là các số...

Ta có ${g(x)=f(x)+f\prime (x)+f^{\prime \prime }(x)=x^3+(3+a) x^2+(b+2 a+6) x+2 a+b+c}$.
Suy ra: ${g\prime (x)=3 x^2+2(3+a) x+b+2 a+6}$.
Xét phương trình
${
\dfrac{f(x)}{g(x)+6}=1 \Leftrightarrow g(x)=f(x)-6 \Leftrightarrow 3 x^2+2(a+3) x+2 a+b+6=0 \Leftrightarrow g\prime (x)=0 \Leftrightarrow\left[\begin{array}{l}
x=x_1 \\
x=x_2
\end{array}\right.
}$
Ta có diện tích bằng
${S=\left|\int_{x_1}^{x_2}\left(\dfrac{f(x)}{g(x)+6}-1\right) {d} x\right|=\left|\int_{x_1}^{x_2}\left(\dfrac{f(x)-g(x)-6}{g(x)+6}\right) {d} x\left\|\int_{x_1}^{x_2}\left(\dfrac{g\prime (x)}{g(x)+6}\right) {d} x|=| \ln |g(x)+6 \|| x_{x_1}^{x_2}\right.\right.}$
${
=|\ln | g\left(x_2\right)+6|-\ln | g\left(x_1\right)+6 \|=|\ln 4|=2 \ln 2
}$