We know that using properties of unit impulse function it can be shown that
$$\int_{t_1}^{t_2} x(t) \delta^{(n)}(t-t_0) dt=(-1)^nx^{(n)}(t_0),\quad t_1<t_0<t_2$$ (source: Continuous and Discrete Signals and Systems- Soliman)
But what would be the case for the problems like these $$\int_{-1}^{2}(3t^3+9)\delta''(t+1) dt$$
I've tried in this way:
$$\int_{-1}^{2}x(t)\delta''(t+1) dt=\\ \big[x(-1)\delta'(t+1)-x'(-1)\delta(t+1)\big]_{-1}^{2}-\int_{-1}^{2}x'(t)\delta'(t+1)dt$$
can't figure out the first part inside the third bracket.