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?

You are correct that this does mean there are no continuous functions satisfying the equation. (There's a quicker way to see this: set $x=1$, then $LHS=0$, but $RHS=5$.)
