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

Options

A. `6 [ (x)/(6) +(x ^((5)/(6)))/( 5 )+ (x ^((2)/(3)))/( 4 ) + (x ^((1)/(2)))/( 3)+ (x ^((1)/(3)))/(2) + x ^((1)/(6))+ log (x ^((1)/(6))-1)]+C`
B. `6 [ (x)/(6) -(x ^((5)/(6)))/( 5 )+ (x ^((2)/(3)))/( 4 ) - (x ^((1)/(2)))/( 3)+ (x ^((1)/(3)))/(2) + x ^((1)/(6))+ log (x ^((1)/(6))-1)]+C`
C. `6 [ (x)/(6) +(x ^((5)/(6)))/( 5 )- (x ^((2)/(3)))/( 4 ) + (x ^((1)/(2)))/( 3)- (x ^((1)/(3)))/(2) + x ^((1)/(6))+ log (x ^((1)/(6))-1)]+C`
D. None of these

Answer

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

### Step 1: Simplify the Integral
First, we rewrite the integral:
\[ \int \frac{x^{1/2}}{x^{1/2} - x^{1/3}} \, dx \]
We can factor out the smallest power in the denominator, which is \( x^{1/3} \):
\[ = \int \frac{x^{1/2}}{x^{1/3}(x^{1/6} - 1)} \, dx \]
This simplifies to:
\[ = \int \frac{x^{1/6}}{x^{1/6} - 1} \, dx \]

### Step 2: Substitution
Let \( t = x^{1/6} \). Then, we differentiate:
\[ dt = \frac{1}{6} x^{-5/6} \, dx \quad \Rightarrow \quad dx = 6 t^5 \, dt \]
Now, substituting \( x = t^6 \) into the integral gives:
\[ = \int \frac{t^3}{t^3 - 1} \cdot 6 t^5 \, dt = 6 \int \frac{t^8}{t^3 - 1} \, dt \]

### Step 3: Long Division
Next, we perform polynomial long division on \( \frac{t^8}{t^3 - 1} \):
1. Divide \( t^8 \) by \( t^3 \) to get \( t^5 \).
2. Multiply \( t^5 \) by \( t^3 - 1 \) to get \( t^8 - t^5 \).
3. Subtract to get \( t^5 \).
4. Divide \( t^5 \) by \( t^3 \) to get \( t^2 \).
5. Continue this process until you reach a remainder of degree less than 3.

The result of the long division will yield:
\[ t^5 + t^2 + t + \frac{1}{t^3 - 1} \]

### Step 4: Integrate Each Term
Now we integrate each term:
\[ 6 \left( \int t^5 \, dt + \int t^2 \, dt + \int t \, dt + \int \frac{1}{t^3 - 1} \, dt \right) \]
Calculating these integrals gives:
\[ = 6 \left( \frac{t^6}{6} + \frac{t^3}{3} + \frac{t^2}{2} + \text{(integral of } \frac{1}{t^3 - 1}) \right) + C \]

### Step 5: Substitute Back
Finally, substitute back \( t = x^{1/6} \):
\[ = x + 2x^{1/2} + 3x^{1/3} + 6 \log(t^3 - 1) + C \]
Thus, the final answer is:
\[ \int \frac{x^{1/2}}{x^{1/2} - x^{1/3}} \, dx = x + 2x^{1/2} + 3x^{1/3} + 6 \log(x^{1/6} - 1) + C \]