Question

\(\int\sqrt{5-2x+x^2}\ dx=\)

• A. \(\frac{x-1}{2}\sqrt{5+2x+x^2}+2\log|(x-1)+\sqrt{5+2x+x^2}|+C\)
• B. \(\frac{x-1}{2}\sqrt{5-2x+x^2}+2\log|(x-1)+\sqrt{x^2+2x+5}|+C\)
• C. \(\frac{x-1}{2}\sqrt{5-2x+x^2}+2\log|(x-1)+\sqrt{5-2x+x^2}|+C\)
• D. \(\frac{x}{2}\sqrt{5-2x+x^2}+4\log|(x-1)+\sqrt{x^2-2x+5}|+C\)

Hint:

When dealing with integrals involving quadratic expressions under square roots, start by completing the square. This allows you to apply standard integral formulas, simplifying the process of solving the integral.

Answer 1

The Correct Option is C

The correct answer is: (C): \(\frac{x-1}{2}\sqrt{5-2x+x^2}+2\log|(x-1)+\sqrt{5-2x+x^2}|+C\)

We are tasked with evaluating the integral:

\(\int \sqrt{5 - 2x + x^2} \, dx\)

Step 1: Complete the square

We begin by completing the square inside the square root. The expression inside the square root is \( 5 - 2x + x^2 \). To complete the square, we rewrite it as:

\( 5 - 2x + x^2 = (x - 1)^2 + 4 \)

Now the integral becomes:

\(\int \sqrt{(x - 1)^2 + 4} \, dx\)

Step 2: Use substitution

We use the substitution \( u = x - 1 \), so that \( du = dx \). The integral becomes:

\(\int \sqrt{u^2 + 4} \, du\)

Step 3: Use standard integral formula

We recognize that this is a standard integral of the form \( \int \sqrt{u^2 + a^2} \, du \), where the solution is:

\( \frac{u}{2}\sqrt{u^2 + a^2} + \frac{a^2}{2} \log|u + \sqrt{u^2 + a^2}| + C \)

Substituting \( a = 2 \), we get:

\( \frac{u}{2}\sqrt{u^2 + 4} + 2 \log|u + \sqrt{u^2 + 4}| + C \)

Step 4: Substitute back and simplify

Substitute \( u = x - 1 \) back into the result to get the final answer:

\( \frac{x - 1}{2}\sqrt{5 - 2x + x^2} + 2 \log|(x - 1) + \sqrt{5 - 2x + x^2}| + C \)

Conclusion:
The correct answer is (C): \(\frac{x - 1}{2}\sqrt{5 - 2x + x^2} + 2 \log|(x - 1) + \sqrt{5 - 2x + x^2}| + C\)