How to find the general antiderivative of $f(x)=x(6-x)^2$?
Sebastian Wright 
I want to find the general antiderivative of $f(x)=x(6-x)^2$. However, I keep getting it wrong.
I am new to antiderivatives, but I think the first thing I should do is differentiate.
$$\frac{d}{dx}(6-x)^2=2(6-x)\cdot-1=-2(6-x)$$
$$f'(x)=[(6-x)^2\cdot1]+[-2(6-x)\cdot x]=(6-x)^2-2x(6-x)$$
According to the antiderivative power rule, when $n \neq -1$, $\int x^n dx= \frac{x^{n+1}}{n+1}+C$.
So it seems like $\int(6-x)^2dx=\frac{(6-x)^3}{3}+C$
However, I can't find a rule that seems like it would work with $-2x(6-x)$. The closest thing that I can find is the "Multiplication by Constant Rule", $\int cf(x)dx=c\cdot\int (f(x))dx$, but I'm not sure if I'm allowed to change $2x(6-x)$ into the form $2(6x-x^2)$.
$$\int2(6x-x^2)$$$$=2 \cdot \int(6x-x^2)$$$$=2 ( 6\int x - \int x^2)$$$$=2 (\frac{6x^2}{2}-\frac{x^3}{3})+C$$
But $\frac{(6-x)^3}{3}- (6x^2-\frac{2x^3}{3})+C$ is incorrect.
$\endgroup$ 13 Answers
$\begingroup$If you differentiate, first, then taking the antiderivative will simply give you the initial function, plus a constant. Instead, we simply integrate. Since $$x(6-x)^2=x(36-12x+x^2)=36x-12x^2+x^3,$$ then $$\int x(6-x)^2\,dx=\int(36x-12x^2+x^3)\,dx=36\int x\,dx-12\int x^2\,dx+\int x^3\,dx,$$ at which point we can use the power rule to get $$\int x(6-x)^2\,dx=18x^2-4x^3+\frac14x^4+C.$$
$\endgroup$ 2 $\begingroup$You can find a primitive in a more simple way:
$$x(x-6)^2=(x-6+6)(x-6)^2=(x-6)^3+6(x-6)^2$$The primitive is therefore: $$F(x)=\frac14(x-6)^4+2(x-6)^3+C$$
$\endgroup$ $\begingroup$Another way to do it is to substitute $y=x-6$, so $$f(x)=6y^2+y^3=\frac{d}{dy}\left(2y^3+\frac{1}{4}y^4\right)=\frac{d}{dx}\frac{y^3(y+8)}{4}=\frac{d}{dx}\frac{(x-6)^3(x+2)}{4}.$$Thus the general antiderivative is $\frac{(x-6)^3(x+2)}{4}+C$.
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