isn't this easy !!

Relevant wiki: Beta Function

$\begin{aligned} I & = \int_0^1 4x^3\left(\frac {d^2}{dx^2} \left(1-x^2\right)^5\right) dx & \small \color{#3D99F6} \text{Let }x = \sin \theta \implies dx = \cos \theta \ d \theta \\ & = \int_0^1 4\sin^3 \theta \cdot \frac {d^2}{dx^2} \cos^{10} \theta \cdot dx \\ & = 4 \int_0^\frac \pi 2 \sin^3 \theta \cdot \frac d{dx} \left(- 10 \cos^8 \theta \sin \theta \right) \cdot \cos \theta \ d \theta \\ & = 40 \int_0^\frac \pi 2 \sin^3 \theta \cos \theta \left(8 \cos^6 \theta \sin^2 \theta - \cos^8 \theta \right) \ d \theta \\ & = 40 \int_0^\frac \pi 2 \left(8 \sin^5 \cos^7 \theta - \sin^3 \theta \cos^9 \theta \right) \ d \theta \\ & = 20 \left(8 B(3,4) - B(2,5) \right) & \small \color{#3D99F6} \text{where } B(m,n) \text{ is the beta function.} \\ & = 20 \left(\frac {8\Gamma (3) \Gamma (4)}{\Gamma (7)} - \frac {\Gamma (2) \Gamma (5)}{\Gamma (7)} \right) & \small \color{#3D99F6} \text{where } \Gamma (z) \text{ is the gamma function.} \\ & = 20 \left(\frac {8 (2!)(3!) - (1!)(4!)}{6!} \right) \\ & = 20 \left(\frac {96 - 24}{720} \right) \\ & = \boxed{2} \end{aligned}$

Let

f(x)=(1-x^2)^5,\\ \begin{aligned} \int_0^1 {4x^3f''(x)dx}&=4x^3f'(x)|_0^1-\int_0^1{12x^2f'(x)dx}\\ &=-\int_0^1{12x^2f'(x)dx}\\ &=\int_0^1{24xf(x)dx}-12x^2f(x)|_0^1\\ &=12\int_0^1{(1-x^2)^5}d(x^2)\\ &=12\int_0^1{u^5}du\\ &=12\times \frac{1}{6}\\ &=2

Differentiating one time and using by parts

The answer is quite straightforward.

$\frac{d^2}{dx^2} (1-x^2)^5 =\frac{d}{dx} 5(1-x^2)^4(-2x)=- 10(x^2 - 1)^4 - 80x^2(x^2 - 1)^3$

This implies the relevant integral is

$\int_0^1 (-40x^3(x^2-1)^4-320x^5(x^2-1)^3) dx$

Let $y=x^2$ , then $dy=2x dx$ , this converts the integral to

$\int_0^1 (-20y(y-1)^4-160y^2(y-1)^3) dy = \int_0^1 (- 180y^5 + 560y^4 - 600y^3 + 240y^2 - 20y) dy$ $= \left.(- 30y^6 + 112y^5 - 150y^4 + 80y^3 - 10y^2)\right|_0^1=-30+112=150+80-10=\boxed{2}$

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