What is the $\int_{-1}^{1} (4t^4 - 4t^2 - 1) e^{-{(t^2 -1)}^2}dt$?

Question

I tried doing this by substitution

$-((t^2 -1)^2) = u$

So $-4t(t^2 -1)dt = du$

but I don't know what to do with the "-1" in $(4t^4 - 4t^2 -1) = (4t^2(t^2 -1) -1)$

Answer 1

HINT 1:

$$\dfrac{d(te^{f(t)})}{dt}=e^{f(t)}(1+tf'(t))$$

HINT 2:

$$\dfrac{d(t^2-1)^2}{dt}=(4t^3-4t)$$

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We have the integral $$\int_{-1}^{1}(4t^4-4t^2-1)e^{-(t^2-1)^2}\,dt=\int_{-1}^{1}(4t^4-4t^2)e^{-(t^2-1)^2}\,dt-\int_{-1}^{1} e^{-(t^2-1)^2}\,dt \tag 1$$For the first integral in $(1)$ we integrate by parts letting $u=-t$ and $v=e^{-(t^2-1)^2}$. Then, integrating by parts reveals that $$\int_{-1}^{1}(4t^4-4t^2)e^{-(t^2-1)^2}\,dt=\left.\left(-te^{-(t-1)^2}\right)\right|_{-1}^{1}+\int_{-1}^{1}e^{-(t-1)^2}\,dt \tag 2$$Substituting the right-hand of $(2)$ into the right-hand side $(1)$, we obtain $$\int_{-1}^{1}(4t^4-4t^2-1)e^{-(t^2-1)^2}\,dt=\left.\left(-te^{-(t-1)^2}\right)\right|_{-1}^{1}=-2$$