If `int(x^(3)dx)/(sqrt(1+x^(2))) = a(1+x^(2))^(3//2) + bsqrt(1+x^(2)) + C`, then

Options

A. `a = (1)/(3), b = 1`
B. `a = (-1)/(3), b = 1`
C. `a = (-1)/(3), b = 1`
D. `a = (1)/(3), b = -1`

Answer

To solve the integral

\[ \int \frac{x^3 \, dx}{\sqrt{1+x^2}}, \]

we will use the substitution method. Let's go through the steps systematically.

### Step 1: Substitution
Let \( t = \sqrt{1+x^2} \). Then, we can express \( x^2 \) in terms of \( t \):
\[ t^2 = 1 + x^2 \implies x^2 = t^2 - 1. \]
Now, differentiate both sides to find \( dx \):
\[ \frac{d}{dx}(t^2) = 2t \frac{dt}{dx} = 2x \implies \frac{dt}{dx} = \frac{x}{t} \implies dx = \frac{t}{x} dt. \]

### Step 2: Express \( x \) in terms of \( t \)
From our substitution, we have:
\[ x = \sqrt{t^2 - 1}. \]
Thus, we can express \( dx \) as:
\[ dx = \frac{t}{\sqrt{t^2 - 1}} dt. \]

### Step 3: Substitute in the integral
Now substitute \( x \) and \( dx \) into the integral:
\[ \int \frac{x^3 \, dx}{\sqrt{1+x^2}} = \int \frac{(\sqrt{t^2 - 1})^3 \cdot \frac{t}{\sqrt{t^2 - 1}} \, dt}{t} = \int \frac{(t^2 - 1)^{3/2} \cdot t}{t \cdot \sqrt{t^2}} \, dt. \]
This simplifies to:
\[ \int (t^2 - 1)^{3/2} dt. \]

### Step 4: Integrate
Now we can integrate:
\[ \int (t^2 - 1)^{3/2} dt. \]
Using the formula for integration, we can find:
\[ \frac{t^4}{4} - t + C. \]

### Step 5: Substitute back to \( x \)
Now we substitute back \( t = \sqrt{1+x^2} \):
\[ = \frac{(1+x^2)^2}{4} - \sqrt{1+x^2} + C. \]

### Step 6: Rearranging the expression
We can express this in the required form:
\[ = \frac{1}{4}(1+x^2)^{3/2} - (1+x^2)^{1/2} + C. \]

### Step 7: Identify coefficients
From the expression, we can identify:
- \( a = \frac{1}{4} \)
- \( b = -1 \)

Thus, the final answer is:
\[ a = \frac{1}{4}, \quad b = -1. \]