Impossible integral? 
$$\int_{0}^{x^2-1} f(t) dt= x^6+x^4+3x^2$$

I saw this problem in a calculus exam. $f$ is assumed to be continuous. Using the Fundamental Theorem of Calculus, I calculated $f(t)= 3t^2+8t+8 $. But when I integrate $f$ it gives me $\int_{0}^{x^2-1} f(t) dt= x^6+x^4+3x^2-5$:
$$\frac {d}{dx}\int_{0}^{x^2-1} f(t) dt= \frac {d}{dx}[x^6+x^4+3x^2]$$
$$f(x^2-1)2x = 6x^5+4x^3+6x$$
$$f(x^2-1) = 3x^4+2x^2+3$$
$$f(x^2-1) = 3(x^2-1)^2+8(x^2-1)+8$$
$$f(t) = 3t^2+8t+8$$
$$\int_{0}^{x^2-1} 3t^2+8t+8 dt= x^6+x^4+3x^2-5$$
This means that there is no continuous function that satisfies the equation of the problem?

Suppose $f(t)=at^2+bt+c$
we have
$$\int_0^{x^2-1} \left(a t^2+b t+c\right) \, dt=$$
$$=x^6+x^4+3 x^2\equiv \frac{a x^6}{3}+x^4 \left(\frac{b}{2}-a\right)+x^2 (a-b+c)-\frac{a}{3}+\frac{b}{2}-c$$
and
$$
\begin{cases}
\frac{a }{3}=1\\
\frac{b}{2}-a=1\\
a-b+c=3\\
-\frac{a}{3}+\frac{b}{2}-c=0\\
\end{cases}
$$
the system has no solution, therefore there is no polynomial $f(t)=at^2+bt+c$ which satisfies the equation.
