If `int frac{xdx}{x^2-4x+8}=K log (x^2-4x+8)+tan^(-1)(frac{x-2}{2})+C`, then the value of K is:

Options

A. 1//2
B. 1
C. 2
D. None of these

Answer

To solve the integral \( \int \frac{x \, dx}{x^2 - 4x + 8} \) and find the value of \( K \) in the expression

\[ \int \frac{x \, dx}{x^2 - 4x + 8} = K \log(x^2 - 4x + 8) + \tan^{-1}\left(\frac{x-2}{2}\right) + C, \]

we will proceed step by step.

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

\[ I = \int \frac{x \, dx}{x^2 - 4x + 8}. \]

### Step 2: Simplify the Denominator
First, we can complete the square for the quadratic in the denominator:

\[ x^2 - 4x + 8 = (x-2)^2 + 4. \]

### Step 3: Use Substitution
Next, we can use the substitution \( u = x^2 - 4x + 8 \). Then, we need to find \( du \):

\[ du = (2x - 4) \, dx \implies dx = \frac{du}{2x - 4}. \]

### Step 4: Express \( x \) in terms of \( u \)
From our substitution, we can express \( x \) in terms of \( u \):

\[ x = 2 + \sqrt{u - 4} \quad \text{or} \quad x = 2 - \sqrt{u - 4}. \]

### Step 5: Rewrite the Integral
Now, substituting back into the integral, we have:

\[ I = \int \frac{x}{u} \cdot \frac{du}{2x - 4}. \]

### Step 6: Split the Integral
We can split the integral into two parts:

\[ I = \frac{1}{2} \int \frac{2x - 4 + 4}{u} \, du = \frac{1}{2} \int \frac{2x - 4}{u} \, du + 2 \int \frac{1}{u} \, du. \]

### Step 7: Evaluate the Integrals
The first integral simplifies to \( \tan^{-1}\left(\frac{x-2}{2}\right) \) and the second integral gives us \( \log|u| \).

### Step 8: Combine Results
Combining the results, we have:

\[ I = \frac{1}{2} \log(x^2 - 4x + 8) + \tan^{-1}\left(\frac{x-2}{2}\right) + C. \]

### Step 9: Compare with Given Expression
Now, comparing this with the given expression:

\[ K \log(x^2 - 4x + 8) + \tan^{-1}\left(\frac{x-2}{2}\right) + C, \]

we see that:

\[ K = \frac{1}{2}. \]

### Final Answer
Thus, the value of \( K \) is:

\[ \boxed{\frac{1}{2}}. \]