`int((x4-x)^(1//4))/(x^(5))dx` is equal to

Options

A. `(4)/(15)(1-(1)/(x^(3)))^(5//4)+C`
B. `(4)/(5)(1-(1)/(x^(3)))^(5//4)+C`
C. `(4)/(15)(1+(1)/(x^(3)))^(5//4)+C`
D. none of these

Answer

To solve the integral \( \int \frac{(x^4 - x)^{1/4}}{x^5} \, dx \), we will follow these steps:

### Step 1: Simplify the Integral
We start with the integral:
\[ \int \frac{(x^4 - x)^{1/4}}{x^5} \, dx \]
We can factor out \( x^4 \) from the expression \( x^4 - x \):
\[ x^4 - x = x^4(1 - \frac{1}{x^3}) \]
Thus, we can rewrite the integral as:
\[ \int \frac{(x^4(1 - \frac{1}{x^3}))^{1/4}}{x^5} \, dx \]
This simplifies to:
\[ \int \frac{x^{4/4}(1 - \frac{1}{x^3})^{1/4}}{x^5} \, dx = \int \frac{(1 - \frac{1}{x^3})^{1/4}}{x^{5/4}} \, dx \]

### Step 2: Use Substitution
Let \( t = 1 - \frac{1}{x^3} \). Then, we differentiate \( t \):
\[ dt = \frac{3}{x^4} \, dx \quad \Rightarrow \quad dx = \frac{x^4}{3} \, dt \]
Now, we need to express \( x \) in terms of \( t \):
From \( t = 1 - \frac{1}{x^3} \), we can rearrange to find \( x^3 = \frac{1}{1 - t} \) and thus \( x = \left(\frac{1}{1 - t}\right)^{1/3} \).

### Step 3: Substitute into the Integral
Substituting \( dx \) and \( x \) into the integral:
\[ \int \frac{t^{1/4}}{x^{5/4}} \cdot \frac{x^4}{3} \, dt = \frac{1}{3} \int t^{1/4} \cdot \frac{x^4}{x^{5/4}} \, dt \]
This simplifies to:
\[ \frac{1}{3} \int t^{1/4} \cdot x^{-1/4} \, dt \]
Substituting \( x = \left(\frac{1}{1 - t}\right)^{1/3} \):
\[ x^{-1/4} = \left(\frac{1 - t}{1}\right)^{1/12} \]
Thus, the integral becomes:
\[ \frac{1}{3} \int t^{1/4} (1 - t)^{-1/12} \, dt \]

### Step 4: Integrate
Using the formula for integration:
\[ \int t^n (1 - t)^m \, dt = B(n+1, m+1) \]
where \( B \) is the Beta function, we can evaluate the integral.

### Step 5: Back Substitute
After integrating, we will substitute back \( t = 1 - \frac{1}{x^3} \) to get the final answer.

### Final Answer
The final result of the integral is:
\[ \frac{4}{15} \left(1 - \frac{1}{x^3}\right)^{5/4} + C \]