`int(x^(3))/((1+x^(2))^(1//3))dx` is equal to

Options

A. `(20)/(3)(1+x^(2))^(2//3)(2x^(2)-3)+C`
B. `(3)/(20)(1+x^(2))^(2//3)(2x^(2)-3)+C`
C. `(3)/(20)(1+x^(2))^(2//3)(2x^(2)+3)+C`
D. none of these

Answer

To solve the integral \( \int \frac{x^3}{(1+x^2)^{1/3}} \, dx \), we will use a substitution method. Here are the steps:

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

### Step 2: Express \( x^3 \) in terms of \( t \)
We can express \( x^3 \) as:
\[ x^3 = x \cdot x^2 = x(t - 1) \]
Substituting \( x = \sqrt{t - 1} \) (since \( x^2 = t - 1 \)):
\[ x^3 = \sqrt{t - 1} \cdot (t - 1) = (t - 1)^{3/2} \]

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

### Step 4: Simplify the integrand
We can rewrite the integrand:
\[ \frac{(t - 1)^{3/2}}{t^{1/3}} = (t - 1)^{3/2} \cdot t^{-1/3} \]
Now we can expand \( (t - 1)^{3/2} \) using the binomial theorem or directly integrate.

### Step 5: Integrate
Using the binomial expansion:
\[ (t - 1)^{3/2} = t^{3/2} - \frac{3}{2}t^{1/2} + \frac{3}{8}t^{-1/2} \]
Thus:
\[ \int (t - 1)^{3/2} t^{-1/3} \, dt = \int \left( t^{3/2 - 1/3} - \frac{3}{2} t^{1/2 - 1/3} + \frac{3}{8} t^{-1/2 - 1/3} \right) dt \]
Calculating the powers:
- \( 3/2 - 1/3 = \frac{9}{6} - \frac{2}{6} = \frac{7}{6} \)
- \( 1/2 - 1/3 = \frac{3}{6} - \frac{2}{6} = \frac{1}{6} \)
- \( -1/2 - 1/3 = -\frac{3}{6} - \frac{2}{6} = -\frac{5}{6} \)

So we have:
\[ \int \left( t^{7/6} - \frac{3}{2} t^{1/6} + \frac{3}{8} t^{-5/6} \right) dt \]

### Step 6: Integrate term by term
Integrating each term:
\[ \int t^{7/6} \, dt = \frac{t^{13/6}}{13/6} = \frac{6}{13} t^{13/6} \]
\[ \int t^{1/6} \, dt = \frac{t^{7/6}}{7/6} = \frac{6}{7} t^{7/6} \]
\[ \int t^{-5/6} \, dt = \frac{t^{1/6}}{1/6} = 6 t^{1/6} \]

### Step 7: Combine results
Putting it all together:
\[ \frac{1}{2} \left( \frac{6}{13} t^{13/6} - \frac{9}{7} t^{7/6} + \frac{9}{4} t^{1/6} \right) + C \]

### Step 8: Substitute back \( t = 1 + x^2 \)
Finally, substituting back \( t = 1 + x^2 \):
\[ = \frac{3}{13} (1 + x^2)^{13/6} - \frac{9}{14} (1 + x^2)^{7/6} + \frac{9}{8} (1 + x^2)^{1/6} + C \]

### Final Answer
Thus, the integral \( \int \frac{x^3}{(1+x^2)^{1/3}} \, dx \) is equal to:
\[ \frac{3}{13} (1 + x^2)^{13/6} - \frac{9}{14} (1 + x^2)^{7/6} + \frac{9}{8} (1 + x^2)^{1/6} + C \]