`int (x^(2))/(sqrt(x^(2)+2x^(3)+3))dx=?`

Options

A. `(1)/(3)log |(x^(3)+1)+sqrt(x^(6)+2x^(3)+3)|+C`
B. ` log |x^(3)+sqrt(x^(6)+2x^(3)+3)|+C`
C. `(1)/(3) log |(x^(3)+1)-sqrt(x^(6)+2x^(3)+3)|+C`
D. None of these

Explanation

To solve the integral \( \int \frac{x^2}{\sqrt{x^6 + 2x^3 + 3}} \, dx \), we will follow these steps:

### Step 1: Simplify the expression under the square root
First, let's rewrite the expression under the square root:
\[ x^6 + 2x^3 + 3 \]
This expression can be analyzed for simplification, but for now, we will proceed with the integration as it is.

### Step 2: Factor out \( x^3 \) from the square root
Notice that we can factor \( x^3 \) from the terms inside the square root:
\[ \sqrt{x^6 + 2x^3 + 3} = \sqrt{x^3(x^3 + 2) + 3} \]
This does not simplify the integral directly, so we will keep the original expression for now.

### Step 3: Use substitution
Let’s use a substitution to simplify the integral. We can let:
\[ u = x^3 + 1 \quad \Rightarrow \quad du = 3x^2 \, dx \quad \Rightarrow \quad dx = \frac{du}{3x^2} \]
Now, we need to express \( x^2 \) in terms of \( u \):
\[ x^3 = u - 1 \quad \Rightarrow \quad x = (u - 1)^{1/3} \]
Thus,
\[ x^2 = ((u - 1)^{1/3})^2 = (u - 1)^{2/3} \]

### Step 4: Substitute in the integral
Now we substitute \( x^2 \) and \( dx \) into the integral:
\[ \int \frac{(u - 1)^{2/3}}{\sqrt{(u - 1)^2 + 2(u - 1) + 3}} \cdot \frac{du}{3(u - 1)^{2/3}} \]
This simplifies to:
\[ \frac{1}{3} \int \frac{1}{\sqrt{(u - 1)^2 + 2(u - 1) + 3}} \, du \]

### Step 5: Simplify the expression under the square root
Now, simplify the expression inside the square root:
\[ (u - 1)^2 + 2(u - 1) + 3 = u^2 - 2u + 1 + 2u - 2 + 3 = u^2 + 2 \]
Thus, the integral becomes:
\[ \frac{1}{3} \int \frac{1}{\sqrt{u^2 + 2}} \, du \]

### Step 6: Solve the integral
The integral \( \int \frac{1}{\sqrt{u^2 + 2}} \, du \) can be solved using the formula:
\[ \int \frac{1}{\sqrt{x^2 + a^2}} \, dx = \ln |x + \sqrt{x^2 + a^2}| + C \]
In our case, \( a^2 = 2 \), so we have:
\[ \int \frac{1}{\sqrt{u^2 + 2}} \, du = \ln |u + \sqrt{u^2 + 2}| + C \]

### Step 7: Substitute back
Now substitute back for \( u \):
\[ = \ln |(x^3 + 1) + \sqrt{(x^3 + 1)^2 + 2}| + C \]

### Step 8: Final expression
Thus, the final result of the integral is:
\[ \frac{1}{3} \ln |(x^3 + 1) + \sqrt{(x^3 + 1)^2 + 2}| + C \]