Question: $\int_{-2}^3f(x^3-2x^2-5x+6)dx$
1. **State the problem.**
We want to evaluate the integral $\int_{-2}^3 f(x^3-2x^2-5x+6)\,dx$.
2. **Factor the inside expression.**
First, factor $x^3-2x^2-5x+6$.
Let $p(x)=x^3-2x^2-5x+6$.
Test simple rational roots:
$p(1)=1-2-5+6=0$.
So $x-1$ is a factor.
Divide to get:
$$x^3-2x^2-5x+6=(x-1)(x^2-x-6).$$
Then factor the quadratic:
$$x^2-x-6=(x-3)(x+2).$$
So
$$x^3-2x^2-5x+6=(x+2)(x-1)(x-3).$$
3. **Use the roots to split the interval.**
The inside expression changes sign at $x=-2$, $x=1$, and $x=3$.
So on $[-2,3]$, the substitution $u=x^3-2x^2-5x+6$ is useful only if we know how the interval maps.
4. **Find the derivative.**
$$\frac{du}{dx}=3x^2-4x-5.$$
This does not match the factorization directly, so we cannot treat it as a simple $u$-substitution by itself.
5. **Notice the problem is incomplete without more information about $f$.**
Because $f$ is an arbitrary function, the value of
$$\int_{-2}^3 f(x^3-2x^2-5x+6)\,dx$$
cannot be determined further from the expression alone.
6. **Final answer.**
The integral cannot be evaluated uniquely unless more is given about $f$.