If `int(x^(5))/(sqrt(1+x^(3)))dx` is equal to

Options

A. `(2)/(9)(1+x^(3))^(5//2)+(2)/(3)(1+x^(3))^(3//2)+C`
B. `(2)/(9)(1+x^(3))^(3//2)-(2)/(3)(1+x^(3))^(1//2)+C`
C. `log|sqrt(x)+sqrt(1+x^(3))|+C`
D. `x^(2)log(1+x^(3))+C`

Answer

To solve the integral \( \int \frac{x^5}{\sqrt{1+x^3}} \, dx \), we can follow these steps:

### Step 1: Rewrite the Integral
We can express \( x^5 \) as \( x^2 \cdot x^3 \):
\[ \int \frac{x^5}{\sqrt{1+x^3}} \, dx = \int \frac{x^2 \cdot x^3}{\sqrt{1+x^3}} \, dx \]

### Step 2: Substitution
Let \( t^2 = 1 + x^3 \). Then, differentiating both sides gives:
\[ 2t \, dt = 3x^2 \, dx \quad \Rightarrow \quad dx = \frac{2t}{3x^2} \, dt \]

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

### Step 4: Substitute Back into the Integral
Now, substituting \( x^3 \) and \( dx \) into the integral:
\[ \int \frac{x^2 (t^2 - 1)}{\sqrt{t^2}} \cdot \frac{2t}{3x^2} \, dt \]
The \( x^2 \) terms cancel out:
\[ = \int \frac{(t^2 - 1) \cdot 2t}{3t} \, dt = \frac{2}{3} \int (t^2 - 1) \, dt \]

### Step 5: Integrate
Now we can integrate:
\[ \frac{2}{3} \left( \int t^2 \, dt - \int 1 \, dt \right) = \frac{2}{3} \left( \frac{t^3}{3} - t \right) + C \]
\[ = \frac{2}{9} t^3 - \frac{2}{3} t + C \]

### Step 6: Substitute Back for \( t \)
Now, substitute back \( t = \sqrt{1 + x^3} \):
\[ = \frac{2}{9} (1 + x^3)^{3/2} - \frac{2}{3} \sqrt{1 + x^3} + C \]

### Final Result
Thus, the final result of the integral is:
\[ \int \frac{x^5}{\sqrt{1+x^3}} \, dx = \frac{2}{9} (1 + x^3)^{3/2} - \frac{2}{3} \sqrt{1 + x^3} + C \]