The value of ∫104x³d²dx²(1−x²)⁵dx is?

Concept: Integration by parts
∫f(x)g(x)dx = f(x)∫g(x)dx - ∫(d/dx[f(x)]∫g(x)dx)dx
∫104x³d²/dx²(1-x²)⁵dx = [4x³d/dx(1-x²)⁵]10 - ∫1012x²d/dx(1-x²)⁵dx
= [4x³ × 5(1-x²)⁴(-2x)]10 - ∫1012x²(-10x(1-x²)⁴)dx
= [4x³ × -10x(1-x²)⁴]10 + ∫10120x³(1-x²)⁴dx
= 0 + 120∫10x³(1-x²)⁴dx
Let u = 1-x² => du = -2xdx
∫x³(1-x²)⁴dx = ∫(1-u)⁴/2(-du)
= 1/2∫(u⁴ - 4u³ + 6u² - 4u + 1)du
= 1/2[u⁵/5 - u⁴ + 2u³ - 2u² + u] + C
= 1/2[(1-x²)⁵/5 - (1-x²)⁴ + 2(1-x²)³ - 2(1-x²)² + (1-x²)] + C
∫10120x³(1-x²)⁴dx = 120/2[(1-x²)⁵/5 - (1-x²)⁴ + 2(1-x²)³ - 2(1-x²)² + (1-x²)]10
= 60[(0 - 0 + 0 - 0 + 0) - (1/5 - 1 + 2 - 2 + 1)]
= 60(-1/5) = -12
Therefore, the value of the integral is -12.