`int(dx)/(x^(2)-2x+2)=`

Options

A. `1/2tan^(-1)(x-1)+c`
B. `tan^(-1)(x-1)+c`
C. `1/2logabs((x+sqrt(2))/(x-sqrt(2)))+c`
D. `1/2logabs((x-sqrt(2))/(x+sqrt(2)))+c`

Answer

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

### Step 1: Rewrite the denominator
First, we need to rewrite the quadratic expression in the denominator. We can complete the square for \(x^2 - 2x + 2\).

\[ x^2 - 2x + 2 = (x^2 - 2x + 1) + 1 = (x - 1)^2 + 1 \]

### Step 2: Substitute the expression in the integral
Now, we can substitute this back into the integral:

\[ \int \frac{dx}{(x - 1)^2 + 1} \]

### Step 3: Recognize the standard form
The integral \(\int \frac{dx}{a^2 + x^2}\) has a standard result, which is \(\frac{1}{a} \tan^{-1} \left(\frac{x}{a}\right) + C\). In our case, \(a = 1\) and we have \(x - 1\) in place of \(x\).

### Step 4: Apply the formula
Using the formula, we can now integrate:

\[ \int \frac{dx}{(x - 1)^2 + 1} = \tan^{-1}(x - 1) + C \]

### Final Answer
Thus, the final answer for the integral is:

\[ \int \frac{dx}{x^2 - 2x + 2} = \tan^{-1}(x - 1) + C \]

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