`int(x^3dx)/(sqrt(1+x^2))i se q u a lto` `1/3sqrt(1+x^2)(2+x^2)+C` `1/3sqrt(1+x^2)(x^2-1)+C` `1/3(1+x^2)^(3/2)+C` (d) `1/3sqrt(1+x^2)(x^2-2)+C`

Options

A. `(1)/(3)sqrt(1+x^(2))(2+x^(2))+C`
B. `(1)/(3)sqrt(1+x^(2))(x^(2)-1)+C`
C. `(1)/(3)(1+x^(2))^(3//2)+C`
D. `(1)/(3)sqrt(1+x^(2))(x^(2)-2)+C`

Explanation

To solve the integral \( \int \frac{x^3}{\sqrt{1+x^2}} \, dx \), we can follow these steps:

### Step 1: Substitute \( t = \sqrt{1+x^2} \)
Let \( t = \sqrt{1+x^2} \). Then, squaring both sides gives us:
\[ t^2 = 1 + x^2 \implies x^2 = t^2 - 1 \]
Next, we differentiate \( t \) with respect to \( x \):
\[ \frac{dt}{dx} = \frac{x}{\sqrt{1+x^2}} \implies dt = \frac{x}{\sqrt{1+x^2}} \, dx \implies dx = \frac{\sqrt{1+x^2}}{x} \, dt \]

### Step 2: Rewrite the integral
Now we can express \( x^3 \) in terms of \( t \):
\[ x^3 = x \cdot x^2 = x(t^2 - 1) \]
Substituting \( dx \) and \( x^3 \) into the integral gives:
\[ \int \frac{x(t^2 - 1)}{t} \cdot \frac{\sqrt{1+x^2}}{x} \, dt = \int (t^2 - 1) \, dt \]

### Step 3: Simplify the integral
The integral simplifies to:
\[ \int (t^2 - 1) \, dt = \int t^2 \, dt - \int 1 \, dt \]
Calculating these integrals:
\[ \int t^2 \, dt = \frac{t^3}{3} \quad \text{and} \quad \int 1 \, dt = t \]
Thus, we have:
\[ \int (t^2 - 1) \, dt = \frac{t^3}{3} - t + C \]

### Step 4: Substitute back for \( t \)
Now we substitute back \( t = \sqrt{1+x^2} \):
\[ \frac{(\sqrt{1+x^2})^3}{3} - \sqrt{1+x^2} + C = \frac{(1+x^2)^{3/2}}{3} - \sqrt{1+x^2} + C \]

### Final Answer
The final answer is:
\[ \frac{(1+x^2)^{3/2}}{3} - \sqrt{1+x^2} + C \]