The value of `int(dx)/(xsqrt(1-x^(3)))` is equal to

Options

A. `1/3ln |(sqrt(1-x^(2))-1)/(sqrt(1-x^(2))+1)|+C`
B. `1/2"ln"|(sqrt(1-x^(2))+1)/(sqrt(1-x^(2))-1)|+C`
C. `1/3ln|1/sqrt(1-x^(3))|+C`
D. `1/3ln|1-x^(3)|+C`

Answer

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

### Step 1: Substitution
We start by substituting \( 1 - x^3 = t^2 \). This gives us:
\[ dx = \frac{d(1 - x^3)}{-3x^2} = -\frac{1}{3x^2} dt \]

### Step 2: Rewrite the Integral
Substituting \( dx \) into the integral, we have:
\[ I = \int \frac{-\frac{1}{3x^2} dt}{x \sqrt{t^2}} = \int \frac{-\frac{1}{3x^2} dt}{x t} \]
Since \( \sqrt{t^2} = t \) (assuming \( t \geq 0 \)), we can rewrite the integral as:
\[ I = -\frac{1}{3} \int \frac{dt}{x^3} \]

### Step 3: Express \( x^3 \) in terms of \( t \)
From our substitution \( 1 - x^3 = t^2 \), we can express \( x^3 \) as:
\[ x^3 = 1 - t^2 \]
Thus, the integral becomes:
\[ I = -\frac{1}{3} \int \frac{dt}{1 - t^2} \]

### Step 4: Integral of \( \frac{1}{1 - t^2} \)
The integral \( \int \frac{dt}{1 - t^2} \) can be solved using the formula:
\[ \int \frac{dt}{1 - t^2} = \frac{1}{2} \ln \left| \frac{1 + t}{1 - t} \right| + C \]
So we have:
\[ I = -\frac{1}{3} \cdot \frac{1}{2} \ln \left| \frac{1 + t}{1 - t} \right| + C = -\frac{1}{6} \ln \left| \frac{1 + t}{1 - t} \right| + C \]

### Step 5: Substitute back for \( t \)
Now we substitute back \( t = \sqrt{1 - x^3} \):
\[ I = -\frac{1}{6} \ln \left| \frac{1 + \sqrt{1 - x^3}}{1 - \sqrt{1 - x^3}} \right| + C \]

### Final Answer
Thus, the value of the integral is:
\[ I = -\frac{1}{6} \ln \left| \frac{1 + \sqrt{1 - x^3}}{1 - \sqrt{1 - x^3}} \right| + C \]