Evaluate :
`int(x+2)/(sqrt(4x-x^(2)))dx`

Answer

To evaluate the integral

\[ \int \frac{x + 2}{\sqrt{4x - x^2}} \, dx, \]

we can follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:

\[ I = \int \frac{x + 2}{\sqrt{4x - x^2}} \, dx. \]

### Step 2: Identify the Denominator
The expression under the square root can be rewritten as:

\[ 4x - x^2 = - (x^2 - 4x) = - (x^2 - 4x + 4 - 4) = - ((x - 2)^2 - 4) = 4 - (x - 2)^2. \]

Thus, we can rewrite the integral as:

\[ I = \int \frac{x + 2}{\sqrt{4 - (x - 2)^2}} \, dx. \]

### Step 3: Use Substitution
Let \( u = x - 2 \), then \( du = dx \) and \( x = u + 2 \). The integral becomes:

\[ I = \int \frac{(u + 2) + 2}{\sqrt{4 - u^2}} \, du = \int \frac{u + 4}{\sqrt{4 - u^2}} \, du. \]

### Step 4: Split the Integral
We can split the integral into two parts:

\[ I = \int \frac{u}{\sqrt{4 - u^2}} \, du + \int \frac{4}{\sqrt{4 - u^2}} \, du. \]

### Step 5: Evaluate the First Integral
For the first integral, we can use the substitution \( v = 4 - u^2 \), thus \( dv = -2u \, du \) or \( du = -\frac{dv}{2u} \). The integral simplifies to:

\[ \int \frac{u}{\sqrt{4 - u^2}} \, du = -\frac{1}{2} \int \sqrt{v} \, dv = -\frac{1}{2} \cdot \frac{2}{3} v^{3/2} = -\frac{1}{3} (4 - u^2)^{3/2}. \]

### Step 6: Evaluate the Second Integral
The second integral is a standard integral:

\[ \int \frac{4}{\sqrt{4 - u^2}} \, du = 4 \sin^{-1} \left( \frac{u}{2} \right). \]

### Step 7: Combine the Results
Combining both integrals, we have:

\[ I = -\frac{1}{3} (4 - u^2)^{3/2} + 4 \sin^{-1} \left( \frac{u}{2} \right) + C. \]

### Step 8: Substitute Back
Now, substituting back \( u = x - 2 \):

\[ I = -\frac{1}{3} (4 - (x - 2)^2)^{3/2} + 4 \sin^{-1} \left( \frac{x - 2}{2} \right) + C. \]

### Final Answer
Thus, the evaluated integral is:

\[ \int \frac{x + 2}{\sqrt{4x - x^2}} \, dx = -\frac{1}{3} (4 - (x - 2)^2)^{3/2} + 4 \sin^{-1} \left( \frac{x - 2}{2} \right) + C. \]

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