$\int\limits_0^1 \underbrace{4 x^3}_{\mathrm{I}} \underbrace{\frac{d^2}{d x^2}\left(1-x^2\right)^5}_{\mathrm{II}} d x$

Integrating by parts:

$$ I=4 x^3\left[\frac{d}{d x}\left(1-x^2\right)^5\right]_0^1-\int\limits_0^1 12 x^2 \frac{d}{d x}\left(1-x^2\right)^5 d x $$

$\begin{aligned} & =4 x^3\left[5\left(1-x^2\right)^4(-2 x)\right]_0^1-12\left[\left[x^2\left(1-x^2\right)^5\right]_0^1-\int\limits_0^1 2 x\left(1-x^2\right)^5 d x\right. \\\\ & =0-0+12 \int\limits_0^1 2 x\left(1-x^2\right)^5 d x\end{aligned}$

Now, putting $1-x^2=t$, we get $-2 x d r=d t$. Therefore,

$$ I=-12 \int\limits_1^0 t^5 d t $$

$\begin{aligned} & \text { When } x=0, t=1 . \\\\ & \text { When } x=1, t=0 .\end{aligned}$

$I=12 \times \int\limits_0^1 t^5 d t=12 \times\left[\frac{t^6}{6}\right]_0^1=12 \times \frac{1}{6}=2$