`int ( x^(2))/(sqrt(1+x^(3)))dx`

Answer

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

### Step 1: Substitution
Let \( t^2 = 1 + x^3 \). This implies that \( x^3 = t^2 - 1 \).

### Step 2: Differentiate to find \( dx \)
Differentiating both sides with respect to \( x \):
\[ \frac{d}{dx}(t^2) = 3x^2 \implies 2t \frac{dt}{dx} = 3x^2 \implies dx = \frac{2t}{3x^2} dt \]

### Step 3: Express \( x^2 \) in terms of \( t \)
From the substitution \( t^2 = 1 + x^3 \), we can express \( x^2 \) as follows:
\[ x^3 = t^2 - 1 \implies x = (t^2 - 1)^{1/3} \]
To find \( x^2 \), we can cube both sides:
\[ x^2 = \left((t^2 - 1)^{1/3}\right)^{2} = (t^2 - 1)^{2/3} \]

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

### Step 5: Integrate
Now we can integrate:
\[ \frac{2}{3} \int dt = \frac{2}{3} t + C \]

### Step 6: Back-substitute for \( t \)
Recall that \( t = \sqrt{1 + x^3} \):
\[ = \frac{2}{3} \sqrt{1 + x^3} + C \]

### Final Answer
Thus, the integral evaluates to:
\[ \int \frac{x^2}{\sqrt{1+x^3}} \, dx = \frac{2}{3} \sqrt{1 + x^3} + C \]

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