Một phương trình bậc ba nghiệm kép:
$(x-1)^2(x+2)=0$
$\to f'(x)=(x-1)^2(x+2)=(x^2-2x+1)(x+2)=x^3-2x^2+x+2x^2-4x+2=x^3-3x+2$
$\to \displaystyle\int f'(x)dx=\displaystyle\int\left( x^3-3x+2\right)dx=\dfrac{1}{4}x^4-\dfrac{3}{2}x^2+2x+C$
Với mỗi giá trị $C$ bất kì ta có một hàm số bậc bốn $f(x)$ cần tìm:
$f(x)=\dfrac{1}{4}x^4-\dfrac{3}{2}x^2+2x$
$f(x)=\dfrac{1}{4}x^4-\dfrac{3}{2}x^2+2x+1$
$f(x)=\dfrac{1}{4}x^4-\dfrac{3}{2}x^2+2x+2$
$f(x)=\dfrac{1}{4}x^4-\dfrac{3}{2}x^2+2x+3$
$f(x)=\dfrac{1}{4}x^4-\dfrac{3}{2}x^2+2x+4$
$f(x)=\dfrac{1}{4}x^4-\dfrac{3}{2}x^2+2x+5$
* Tương tự:
$f'(x)=(x+1)^2(x-1)=(x^2+2x+1)(x-1)=x^3+2x^2+x-x^2-2x-1=x^3+x^2-x-1$
$\to \displaystyle\int f'(x)dx=\dfrac{1}{4}x^4+\dfrac{1}{3}x^3-\dfrac{1}{2}x^2-x+C$
Ta có một số $f(x)$:
$f(x)= \dfrac{1}{4}x^4+\dfrac{1}{3}x^3-\dfrac{1}{2}x^2-x+1$
$f(x)=\dfrac{1}{4}x^4+\dfrac{1}{3}x^3-\dfrac{1}{2}x^2-x+2$
$f(x)=\dfrac{1}{4}x^4+\dfrac{1}{3}x^3-\dfrac{1}{2}x^2-x+3$
$f(x)=\dfrac{1}{4}x^4+\dfrac{1}{3}x^3-\dfrac{1}{2}x^2-x+4$
$\dfrac{1}{4}x^4+\dfrac{1}{3}x^3-\dfrac{1}{2}x^2-x+2021$
