Question

describe a first step in integrating $int\frac{6}{x^{2}-4x + 13}dx$. choose the correct answer below. a. use u - substitution with $u=x^{2}-4x + 13$. b. split the fraction into parts. c. complete the square on $x^{2}-4x + 13$. d. use u - substitution with $u=x^{2}$. e. factor the denominator.

Explanation:

Step1: Analyze integral form

The integral is of the form $\int\frac{6}{x^{2}-4x + 13}dx$. Completing the square in the denominator $x^{2}-4x + 13$ is a common first - step for integrals of this type. The general form for completing the square of $ax^{2}+bx + c$ is $a(x+\frac{b}{2a})^{2}+c-\frac{b^{2}}{4a}$. For $x^{2}-4x + 13$, we have $a = 1$, $b=-4$, $c = 13$. Completing the square gives $(x - 2)^{2}+9$. After completing the square, the integral can be rewritten in a form that may be integrated using standard integral formulas (such as the arctangent integral formula).

Step2: Evaluate options

• Option A: If $u=x^{2}-4x + 13$, then $du=(2x - 4)dx$, and there is no $2x - 4$ in the numerator, so this substitution is not helpful.
• Option B: Splitting the fraction into parts is not applicable as the numerator is a constant and the denominator is a non - factorable quadratic (in real numbers).
• Option C: As explained above, completing the square on $x^{2}-4x + 13$ is a valid first step.
• Option D: If $u=x^{2}$, then $du = 2xdx$, and this substitution does not simplify the integral.
• Option E: The discriminant of $x^{2}-4x + 13$ is $\Delta=b^{2}-4ac=(-4)^{2}-4\times1\times13=16 - 52=-36<0$, so it cannot be factored over the real numbers.

Answer:

C. Complete the square on $x^{2}-4x + 13$