Key Concepts and Formulas

• Completing the square: $x^2 + bx + c = (x + \frac{b}{2})^2 + (c - \frac{b^2}{4})$
• Trigonometric substitution: For integrals involving $a^2 + x^2$, we can use $x = a \tan \theta$.
• Trigonometric identities: $\sin 2\theta = 2 \sin \theta \cos \theta$, $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$, $1 + \tan^2 \theta = \sec^2 \theta$

Step-by-Step Solution

Step 1: Rewrite the integrand by completing the square. We have $x^2 - 2x + 10 = (x^2 - 2x + 1) + 9 = (x-1)^2 + 9$. Therefore, the integral becomes: $I = \int \frac{dx}{((x-1)^2 + 9)^2}$ This step helps us to identify the appropriate trigonometric substitution.

Step 2: Apply the trigonometric substitution $x - 1 = 3 \tan \theta$. Let $x - 1 = 3 \tan \theta$. Then, $dx = 3 \sec^2 \theta \, d\theta$. Substituting these into the integral gives: $I = \int \frac{3 \sec^2 \theta \, d\theta}{((3 \tan \theta)^2 + 9)^2} = \int \frac{3 \sec^2 \theta \, d\theta}{(9 \tan^2 \theta + 9)^2} = \int \frac{3 \sec^2 \theta \, d\theta}{(9(\tan^2 \theta + 1))^2}$ This substitution simplifies the integral into a trigonometric form.

Step 3: Simplify the integrand using the identity $1 + \tan^2 \theta = \sec^2 \theta$. $I = \int \frac{3 \sec^2 \theta \, d\theta}{(9 \sec^2 \theta)^2} = \int \frac{3 \sec^2 \theta \, d\theta}{81 \sec^4 \theta} = \frac{3}{81} \int \frac{d\theta}{\sec^2 \theta} = \frac{1}{27} \int \cos^2 \theta \, d\theta$ This step simplifies the integral further.

Step 4: Use the identity $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$ to rewrite the integrand. $I = \frac{1}{27} \int \frac{1 + \cos 2\theta}{2} \, d\theta = \frac{1}{54} \int (1 + \cos 2\theta) \, d\theta$ This step prepares the integral for direct integration.

Step 5: Integrate with respect to $\theta$. $I = \frac{1}{54} \left( \theta + \frac{1}{2} \sin 2\theta \right) + C = \frac{1}{54} \left( \theta + \sin \theta \cos \theta \right) + C$ This step performs the integration.

Step 6: Convert back to $x$ using the substitution $x - 1 = 3 \tan \theta$. Since $x - 1 = 3 \tan \theta$, we have $\tan \theta = \frac{x-1}{3}$, so $\theta = \tan^{-1} \left( \frac{x-1}{3} \right)$. Also, $\sin \theta = \frac{x-1}{\sqrt{(x-1)^2 + 9}}$ and $\cos \theta = \frac{3}{\sqrt{(x-1)^2 + 9}}$. Therefore, $\sin \theta \cos \theta = \frac{3(x-1)}{(x-1)^2 + 9} = \frac{3(x-1)}{x^2 - 2x + 10}$.

Substituting these back into the integral gives: $I = \frac{1}{54} \left( \tan^{-1} \left( \frac{x-1}{3} \right) + \frac{3(x-1)}{x^2 - 2x + 10} \right) + C$

Step 7: Identify $A$ and $f(x)$. Comparing this with the given form $A \left( \tan^{-1} \left( \frac{x-1}{3} \right) + \frac{f(x)}{x^2 - 2x + 10} \right) + C$, we have $A = \frac{1}{54}$ and $f(x) = 3(x-1)$.

Common Mistakes & Tips

• Remember to convert back to the original variable $x$ after integrating with respect to $\theta$.
• Be careful with trigonometric identities and algebraic manipulations to avoid errors.
• Double-check the final answer by differentiating it to see if it matches the original integrand.

Summary We solved the integral by completing the square, applying a trigonometric substitution, simplifying the integrand using trigonometric identities, integrating, and converting back to the original variable. Comparing the result with the given form, we found $A = \frac{1}{54}$ and $f(x) = 3(x-1)$. This corresponds to option (B). However, the "Correct Answer" is stated to be option (A). Let's carefully re-examine the provided solution to ensure no error has been made, and that the stated "Correct Answer" is indeed correct.

The integral is $\int \frac{dx}{((x-1)^2 + 9)^2}$. Let $x-1 = 3\tan \theta$, so $dx = 3\sec^2 \theta d\theta$. The integral becomes $\int \frac{3\sec^2 \theta d\theta}{(9\tan^2 \theta + 9)^2} = \int \frac{3\sec^2 \theta d\theta}{81 \sec^4 \theta} = \frac{1}{27} \int \cos^2 \theta d\theta = \frac{1}{27} \int \frac{1+\cos 2\theta}{2} d\theta = \frac{1}{54} (\theta + \frac{\sin 2\theta}{2}) + C = \frac{1}{54} (\theta + \sin \theta \cos \theta) + C$. Since $\tan \theta = \frac{x-1}{3}$, then $\theta = \arctan(\frac{x-1}{3})$. Also $\sin \theta = \frac{x-1}{\sqrt{(x-1)^2 + 9}}$ and $\cos \theta = \frac{3}{\sqrt{(x-1)^2 + 9}}$. So $\sin \theta \cos \theta = \frac{3(x-1)}{(x-1)^2 + 9} = \frac{3(x-1)}{x^2 - 2x + 10}$. The integral is $\frac{1}{54} (\arctan(\frac{x-1}{3}) + \frac{3(x-1)}{x^2 - 2x + 10}) + C$. So $A = \frac{1}{54}$ and $f(x) = 3(x-1)$. This corresponds to option (B).

The given answer states option (A) is the correct answer, with $A = \frac{1}{54}$ and $f(x) = 9(x-1)^2$. This does NOT match the derived result.

The final answer is \boxed{A = 1/54, f(x) = 3(x-1)}, which corresponds to option (B).