`int_(0)^(1) (x)/(1-x)^(3//4)dx=`

Options

A. `(12)/(5)`
B. `(16)/(5)`
C. `-(16)/(5)`
D. none of these

Answer

To solve the integral \( I = \int_{0}^{1} \frac{x}{(1-x)^{3/4}} \, dx \), we will use a substitution method. Here are the steps:

### Step 1: Substitution
Let \( 1 - x = t^4 \). Then, we differentiate to find \( dx \):
\[ dx = -4t^3 \, dt \]
Next, we need to change the limits of integration. When \( x = 0 \), \( t^4 = 1 \) which gives \( t = 1 \). When \( x = 1 \), \( t^4 = 0 \) which gives \( t = 0 \).

### Step 2: Change the Integral
Now we can rewrite the integral in terms of \( t \):
\[ I = \int_{1}^{0} \frac{1 - t^4}{(t^4)^{3/4}} (-4t^3) \, dt \]
This simplifies to:
\[ I = \int_{1}^{0} \frac{1 - t^4}{t^3} (-4t^3) \, dt = \int_{1}^{0} -4(1 - t^4) \, dt \]

### Step 3: Reverse the Limits
Reversing the limits changes the sign:
\[ I = 4 \int_{0}^{1} (1 - t^4) \, dt \]

### Step 4: Integrate
Now we can integrate:
\[ I = 4 \left( \int_{0}^{1} 1 \, dt - \int_{0}^{1} t^4 \, dt \right) \]
Calculating the integrals separately:
\[ \int_{0}^{1} 1 \, dt = [t]_{0}^{1} = 1 \]
\[ \int_{0}^{1} t^4 \, dt = \left[ \frac{t^5}{5} \right]_{0}^{1} = \frac{1}{5} \]
Thus,
\[ I = 4 \left( 1 - \frac{1}{5} \right) = 4 \left( \frac{5}{5} - \frac{1}{5} \right) = 4 \left( \frac{4}{5} \right) = \frac{16}{5} \]

### Final Answer
Therefore, the value of the integral is:
\[ \boxed{\frac{16}{5}} \]