`int (dx)/sqrt((2x-x^2)^3`

Answer

To solve the integral \( I = \int \frac{dx}{\sqrt{(2x - x^2)^3}} \), we will follow a systematic approach.

### Step 1: Rewrite the expression under the square root
First, we can rewrite the expression inside the square root:
\[ 2x - x^2 = - (x^2 - 2x) = - (x(x - 2)) \]
Thus, we have:
\[ I = \int \frac{dx}{\sqrt{(-(x(x - 2)))^3}} = \int \frac{dx}{\sqrt{-(x^3(x - 2)^3)}} \]

### Step 2: Factor out \(x^2\)
Next, we factor out \(x^2\) from the expression:
\[ 2x - x^2 = -x^2(x - 2) \]
This gives us:
\[ I = \int \frac{dx}{\sqrt{(-x^2(x - 2))^3}} = \int \frac{dx}{\sqrt{-x^6(x - 2)^3}} = \int \frac{dx}{(-1)^{3/2} x^3 \sqrt{(x - 2)^3}} \]
Since \((-1)^{3/2} = -i\), we can ignore the imaginary unit for now and focus on the real part.

### Step 3: Substitution
Let’s make the substitution:
\[ t = \frac{2}{x} - 1 \implies x = \frac{2}{t + 1} \]
Differentiating gives:
\[ dx = -\frac{2}{(t + 1)^2} dt \]

### Step 4: Change of variables in the integral
Now substitute \(x\) and \(dx\) in the integral:
\[ I = \int \frac{-\frac{2}{(t + 1)^2} dt}{\sqrt{(2(\frac{2}{t + 1}) - (\frac{2}{t + 1})^2)^3}} \]
Calculating the expression inside the square root:
\[ 2(\frac{2}{t + 1}) - (\frac{2}{t + 1})^2 = \frac{4}{t + 1} - \frac{4}{(t + 1)^2} = \frac{4(t + 1) - 4}{(t + 1)^2} = \frac{4t}{(t + 1)^2} \]
Thus:
\[ I = \int \frac{-\frac{2}{(t + 1)^2} dt}{\sqrt{\left(\frac{4t}{(t + 1)^2}\right)^3}} = \int \frac{-\frac{2}{(t + 1)^2} dt}{\sqrt{\frac{64t^3}{(t + 1)^6}}} \]
This simplifies to:
\[ I = \int \frac{-\frac{2}{(t + 1)^2} dt}{\frac{8t^{3/2}}{(t + 1)^3}} = \int \frac{-2(t + 1) dt}{8t^{3/2}} = -\frac{1}{4} \int \frac{(t + 1) dt}{t^{3/2}} \]

### Step 5: Separate the integral
Now we can separate the integral:
\[ I = -\frac{1}{4} \left( \int t^{-1/2} dt + \int t^{-3/2} dt \right) \]

### Step 6: Integrate
Integrating term by term:
\[ \int t^{-1/2} dt = 2t^{1/2}, \quad \int t^{-3/2} dt = -2t^{-1/2} \]
Thus:
\[ I = -\frac{1}{4} \left( 2t^{1/2} - 2t^{-1/2} \right) + C \]
This simplifies to:
\[ I = -\frac{1}{2} t^{1/2} + \frac{1}{2} t^{-1/2} + C \]

### Step 7: Substitute back for \(t\)
Recalling that \(t = \frac{2}{x} - 1\), we substitute back:
\[ I = -\frac{1}{2} \sqrt{\frac{2}{x} - 1} + \frac{1}{2} \frac{1}{\sqrt{\frac{2}{x} - 1}} + C \]

### Final Result
Thus, the final result of the integral is:
\[ I = \frac{1}{2} \left( \frac{1}{\sqrt{\frac{2}{x} - 1}} - \sqrt{\frac{2}{x} - 1} \right) + C \]