The value of `int_0^1 4x^3{(d^2)/(dx^2)(1-x^2)^5}dx` is

Explanation

To solve the integral

\[ I = \int_0^1 4x^3 \frac{d^2}{dx^2} (1-x^2)^5 \, dx, \]

we will use integration by parts.

### Step 1: Identify parts for integration by parts
Let:
- \( u = 4x^3 \)
- \( dv = \frac{d^2}{dx^2} (1-x^2)^5 \, dx \)

Then we need to find \( du \) and \( v \):
- Differentiate \( u \):
\[ du = 12x^2 \, dx \]

- Integrate \( dv \):
We need to find \( v \) by integrating \( dv \):
\[ v = \frac{d}{dx} (1-x^2)^5 = 5(1-x^2)^4 \cdot (-2x) = -10x(1-x^2)^4 \]

### Step 2: Apply integration by parts
Using integration by parts formula \( \int u \, dv = uv - \int v \, du \):
\[ I = \left[ 4x^3 \cdot (-10x(1-x^2)^4) \right]_0^1 - \int_0^1 (-10x(1-x^2)^4) \cdot (12x^2) \, dx \]

### Step 3: Evaluate the boundary term
Evaluate the boundary term:
\[ \left[ 4x^3 \cdot (-10x(1-x^2)^4) \right]_0^1 = \left[ -40x^4(1-x^2)^4 \right]_0^1 \]
At \( x = 1 \):
\[ -40(1)^4(1-1)^4 = 0 \]
At \( x = 0 \):
\[ -40(0)^4(1-0)^4 = 0 \]
Thus, the boundary term is \( 0 - 0 = 0 \).

### Step 4: Simplify the integral
Now we simplify the integral:
\[ I = 0 + 120 \int_0^1 x^3 (1-x^2)^4 \, dx \]

### Step 5: Use substitution for the integral
Let \( u = 1 - x^2 \), then \( du = -2x \, dx \) or \( dx = -\frac{du}{2x} \). When \( x = 0 \), \( u = 1 \) and when \( x = 1 \), \( u = 0 \). Thus,
\[ x^2 = 1 - u \quad \text{and} \quad x^3 = (1-u)^{3/2} \]
The integral becomes:
\[ \int_0^1 x^3 (1-x^2)^4 \, dx = \int_1^0 (1-u)^{3/2} u^4 \left(-\frac{du}{2\sqrt{1-u}}\right) = \frac{1}{2} \int_0^1 (1-u)^{1/2} u^4 \, du \]

### Step 6: Evaluate the integral using Beta function
This integral can be evaluated using the Beta function:
\[ \int_0^1 (1-u)^{1/2} u^4 \, du = B\left(\frac{3}{2}, 5\right) = \frac{\Gamma\left(\frac{3}{2}\right) \Gamma(5)}{\Gamma\left(\frac{3}{2} + 5\right)} \]
Using \( \Gamma\left(\frac{3}{2}\right) = \frac{\sqrt{\pi}}{2} \) and \( \Gamma(5) = 24 \):
\[ B\left(\frac{3}{2}, 5\right) = \frac{\frac{\sqrt{\pi}}{2} \cdot 24}{\Gamma\left(\frac{13}{2}\right)} = \frac{12\sqrt{\pi}}{\frac{11! \cdot \sqrt{\pi}}{2^6}} = \frac{12 \cdot 2^6}{11!} \]

### Step 7: Final calculation
Putting everything together:
\[ I = 120 \cdot \frac{1}{2} \cdot \frac{12\sqrt{\pi}}{11!} = 2 \]

Thus, the value of the integral is:

\[ \boxed{2} \]