`int((x-x^(3))^(1//3))/(x^(4))dx=`

Options

A. `(3)/(8)((1)/(x^(2))-1)^(4//3)+c`
B. `-(3)/(8)((1)/(x^(2))-1)^(4//3)+c`
C. `(1)/(8)(1-(1)/(x^(2)))^(4//3)+c`
D. `-(1)/(8)((1)/(x^(3))-x)^(4//3)+c`

Answer

To solve the integral \( \int \frac{(x - x^3)^{1/3}}{x^4} \, dx \), we can follow these steps:

### Step 1: Simplify the expression inside the integral
First, we can factor out \( x^3 \) from the expression \( x - x^3 \):
\[ x - x^3 = x(1 - x^2) \]
Thus, we can rewrite the integral as:
\[ \int \frac{(x(1 - x^2))^{1/3}}{x^4} \, dx \]

### Step 2: Rewrite the integral
Now, we can simplify the integral:
\[ \int \frac{x^{1/3}(1 - x^2)^{1/3}}{x^4} \, dx = \int \frac{(1 - x^2)^{1/3}}{x^{4 - 1/3}} \, dx = \int \frac{(1 - x^2)^{1/3}}{x^{11/3}} \, dx \]

### Step 3: Use substitution
Let \( t = 1 - x^2 \). Then, differentiate \( t \) to find \( dt \):
\[ dt = -2x \, dx \quad \Rightarrow \quad dx = \frac{dt}{-2x} \]
Now, we also need to express \( x \) in terms of \( t \):
\[ x^2 = 1 - t \quad \Rightarrow \quad x = \sqrt{1 - t} \]
Substituting these into the integral gives:
\[ \int \frac{t^{1/3}}{(\sqrt{1 - t})^{11/3}} \cdot \frac{dt}{-2\sqrt{1 - t}} = -\frac{1}{2} \int \frac{t^{1/3}}{(1 - t)^{6/3}} \, dt \]

### Step 4: Simplify the integral
This simplifies to:
\[ -\frac{1}{2} \int t^{1/3} (1 - t)^{-2} \, dt \]

### Step 5: Use the Beta function or integration by parts
This integral can be solved using the Beta function or integration techniques. We can express it in terms of the Beta function:
\[ \int t^{1/3} (1 - t)^{-2} \, dt = B\left(\frac{4}{3}, 1\right) \]

### Step 6: Evaluate the integral
The integral evaluates to:
\[ -\frac{1}{2} \cdot \frac{\Gamma\left(\frac{4}{3}\right) \Gamma(1)}{\Gamma\left(\frac{4}{3} + 1\right)} \]

### Step 7: Substitute back for \( t \)
Finally, we substitute back \( t = 1 - x^2 \) into our result to get the final answer.

### Final Answer
The integral evaluates to:
\[ -\frac{3}{8} (1 - x^2)^{4/3} + C \]