`int (1)/(4x^(2)-20x+17)dx`

Answer

To solve the integral \( \int \frac{1}{4x^2 - 20x + 17} \, dx \), we will follow these steps:

### Step 1: Rewrite the Integral
Let \( I = \int \frac{1}{4x^2 - 20x + 17} \, dx \).

### Step 2: Factor out the Constant
We can factor out a 4 from the denominator:
\[ I = \frac{1}{4} \int \frac{1}{x^2 - 5x + \frac{17}{4}} \, dx \]

### Step 3: Complete the Square
Next, we will complete the square for the expression \( x^2 - 5x + \frac{17}{4} \):
\[ x^2 - 5x = \left(x - \frac{5}{2}\right)^2 - \frac{25}{4} \]
Thus,
\[ x^2 - 5x + \frac{17}{4} = \left(x - \frac{5}{2}\right)^2 - \frac{25}{4} + \frac{17}{4} = \left(x - \frac{5}{2}\right)^2 - 2 \]

### Step 4: Substitute Back into the Integral
Now, substituting this back into the integral gives us:
\[ I = \frac{1}{4} \int \frac{1}{\left(x - \frac{5}{2}\right)^2 - 2} \, dx \]

### Step 5: Use the Integral Formula
We can use the formula for the integral \( \int \frac{1}{u^2 - a^2} \, du = \frac{1}{2a} \log \left| \frac{u-a}{u+a} \right| + C \). Here, \( u = x - \frac{5}{2} \) and \( a = \sqrt{2} \):
\[ I = \frac{1}{4} \cdot \frac{1}{2\sqrt{2}} \log \left| \frac{x - \frac{5}{2} - \sqrt{2}}{x - \frac{5}{2} + \sqrt{2}} \right| + C \]

### Step 6: Simplify the Expression
This simplifies to:
\[ I = \frac{1}{8\sqrt{2}} \log \left| \frac{2x - 5 - 2\sqrt{2}}{2x - 5 + 2\sqrt{2}} \right| + C \]

### Final Answer
Thus, the integral evaluates to:
\[ \int \frac{1}{4x^2 - 20x + 17} \, dx = \frac{1}{8\sqrt{2}} \log \left| \frac{2x - 5 - 2\sqrt{2}}{2x - 5 + 2\sqrt{2}} \right| + C \]

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