`int_(0)^(2)(x^(3)dx)/((x^(2)+1)^(3//2))`is equal to

Options

A. `(sqrt2-1)^(2)`
B. `((sqrt2-1)^(2))/(2)`
C. `(sqrt2-1)/(2)`
D. None of these

Explanation

To solve the integral \( I = \int_{0}^{2} \frac{x^3}{(x^2 + 1)^{3/2}} \, dx \), we will use a substitution method.

### Step 1: Substitution
Let \( t = x^2 + 1 \). Then, the differential \( dt = 2x \, dx \) or \( dx = \frac{dt}{2x} \).

### Step 2: Change of Limits
When \( x = 0 \):
\[ t = 0^2 + 1 = 1 \]
When \( x = 2 \):
\[ t = 2^2 + 1 = 5 \]
Thus, the limits change from \( x: 0 \to 2 \) to \( t: 1 \to 5 \).

### Step 3: Express \( x \) in terms of \( t \)
From the substitution \( t = x^2 + 1 \), we can express \( x^2 \) as:
\[ x^2 = t - 1 \quad \text{and thus} \quad x = \sqrt{t - 1} \]

### Step 4: Substitute into the Integral
Now substitute \( x \) and \( dx \) into the integral:
\[ I = \int_{1}^{5} \frac{(\sqrt{t - 1})^3}{t^{3/2}} \cdot \frac{dt}{2\sqrt{t - 1}} = \int_{1}^{5} \frac{(t - 1)^{3/2}}{t^{3/2}} \cdot \frac{dt}{2} \]
This simplifies to:
\[ I = \frac{1}{2} \int_{1}^{5} \frac{(t - 1)^{3/2}}{t^{3/2}} \, dt \]

### Step 5: Simplify the Integral
We can rewrite the integrand:
\[ \frac{(t - 1)^{3/2}}{t^{3/2}} = \left(1 - \frac{1}{t}\right)^{3/2} \]
Thus, we have:
\[ I = \frac{1}{2} \int_{1}^{5} (t - 1)^{3/2} t^{-3/2} \, dt \]

### Step 6: Integration
Now we can integrate:
\[ I = \frac{1}{2} \left[ -\frac{2(t-1)^{5/2}}{5t^{3/2}} \right]_{1}^{5} \]

### Step 7: Evaluate the Integral
Now we evaluate the limits:
\[ I = \frac{1}{2} \left[ -\frac{2(5-1)^{5/2}}{5 \cdot 5^{3/2}} + \frac{2(1-1)^{5/2}}{5 \cdot 1^{3/2}} \right] \]
Calculating this gives:
\[ I = \frac{1}{2} \left[ -\frac{2(4)^{5/2}}{5 \cdot 5\sqrt{5}} \right] \]
This simplifies to:
\[ I = \frac{1}{2} \left[ -\frac{128}{25\sqrt{5}} \right] \]

### Final Step: Simplifying the Result
Thus, the final result is:
\[ I = \frac{64}{25\sqrt{5}} \]

### Final Answer:
\[ \int_{0}^{2} \frac{x^3}{(x^2 + 1)^{3/2}} \, dx = \frac{64}{25\sqrt{5}} \]