Question

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

• A. tan^-1(x-1)+C
• B. sin^-1(2x-1)+C
• C. sin^-1(x-1)+C
• D. tan^-1(2x-1)+C
• E. \(\frac{1}{(2x-1)^3}+C\)

Answer 1

The Correct Option is A

Step 1: Complete the square in the denominator: \[ x^2 - 2x + 2 = (x^2 - 2x + 1) + 1 = (x-1)^2 + 1 \]

Step 2: Rewrite the integral: \[ \int \frac{1}{(x-1)^2 + 1} dx \]

Step 3: Recognize the standard integral form: \[ \int \frac{1}{u^2 + a^2} du = \frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right) + C \] Here, \( u = x-1 \) and \( a = 1 \).

Step 4: Apply the formula: \[ \int \frac{1}{(x-1)^2 + 1} dx = \tan^{-1}(x-1) + C \]

Conclusion: The correct answer is \(\boxed{A}\) (\(\tan^{-1}(x-1) + C\)).

Answer 2

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

We need to find the integral of the given function. 

The denominator \( x^2 - 2x + 2 \) is a quadratic expression that can be rewritten as a perfect square.

Step 1: Complete the square for the denominator

The quadratic expression is \( x^2 - 2x + 2 \). To complete the square:

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

So, the integral becomes:

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

Step 2: Use a standard integral formula

The standard integral formula for the expression \( \int \frac{1}{x^2 + a^2} \, dx \) is:

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

In our case, \( a = 1 \), and the integral becomes:

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

Final Answer:

The correct option is:

\[ \boxed{\text{(A) } \tan^{-1}(x - 1) + C} \]