Evaluate the integral by computing the square. Hint: thinks in term of dx/(a^2 + u^2).        2        0     Frac{dx}{x^2 - 2x +2}

The given integral is:

\[
\int_0^2 \frac{dx}{x^2 - 2x + 2}
\]

We will evaluate this integral by simplifying the denominator and then applying a standard form involving \(\frac{dx}{a^2 + u^2}\).

### Step 1: Complete the square
We start by completing the square for the quadratic expression in the denominator:

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

Thus, the integral becomes:

\[
\int_0^2 \frac{dx}{(x - 1)^2 + 1}
\]

### Step 2: Substitution
Let \( u = x - 1 \), which implies \( du = dx \). This changes the limits of integration as follows:

- When \( x = 0 \), \( u = -1 \)
- When \( x = 2 \), \( u = 1 \)

Now, the integral becomes:

\[
\int_{-1}^{1} \frac{du}{u^2 + 1}
\]

### Step 3: Recognize the standard form
The integral \(\int \frac{du}{u^2 + 1}\) is a standard integral, and its result is the arctangent function:

\[
\int \frac{du}{u^2 + 1} = \tan^{-1}(u)
\]

### Step 4: Evaluate the integral
Now, apply the limits of integration:

\[
\left[ \tan^{-1}(u) \right]_{-1}^{1} = \tan^{-1}(1) - \tan^{-1}(-1)
\]

We know that:

\[
\tan^{-1}(1) = \frac{\pi}{4} \quad \text{and} \quad \tan^{-1}(-1) = -\frac{\pi}{4}
\]

Thus, the integral becomes:

\[
\frac{\pi}{4} - \left( -\frac{\pi}{4} \right) = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}
\]

### Final Answer:
The value of the integral is:

\[
\boxed{\frac{\pi}{2}}
\]

Would you like a detailed explanation on any part of the process? Here are some related questions:

1. How does completing the square simplify integrals?
2. What are the properties of the arctangent function?
3. How does substitution affect the limits of integration?
4. What are other examples of integrals of the form \(\frac{dx}{a^2 + u^2}\)?
5. How can we derive the integral of \(\frac{dx}{x^2 + a^2}\) from first principles?

**Tip:** Always check the form of the denominator when dealing with quadratic expressions—it often simplifies the problem significantly.

Evaluate the integral by computing the square. Hint: thinks in term of dx/(a^2 + u^2). \int_0^2 \frac{dx}{x^2 - 2x +2}

This solution demonstrates how to evaluate the integral of \int_0^2 \frac{dx}{x^2 - 2x + 2} by completing the square and applying substitution. The problem includes the use of the standard arctangent formula. Suitable for students studying undergraduate calculus or advanced high school students.

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