The key skill here is completing the square as preparatory algebra for integrating rational functions with quadratic denominators. For the integrand $ \frac{1}{x^2 + 4x + 8} $, the quadratic does not factor over the reals since the discriminant is negative. Completing the square rewrites it as $ (x + 2)^2 + 4 $, which matches the form $ u^2 + a^2 $ with $ a = 2 $. This enables the use of the integral formula $ \int \frac{du}{u^2 + a^2} = \frac{1}{a} \arctan\left( \frac{u}{a} \right) + C $, resulting in $ \frac{1}{2} \arctan\left( \frac{x + 2}{2} \right) + C $. A tempting distractor is $ \ln(x^2 + 4x + 8) + C $, but it fails because logarithmic antiderivatives apply to factorable quadratics, not this irreducible case. Always compute the discriminant of the quadratic denominator first; if negative, complete the square and apply the arctangent formula for accurate integration.

This problem requires performing polynomial long division as preparatory algebra before integrating the rational function. The algebraic step is necessary because the degree of the numerator is equal to the degree of the denominator plus one, making the rational function improper. To integrate, we must first divide to express it as a polynomial plus a proper fraction. This allows us to integrate the polynomial terms directly and use logarithmic integration for the fractional remainder. A tempting distractor like choice C fails because it omits the linear term from the quotient, leading to an incomplete antiderivative. Always perform long division or simplify rational functions before integration to ensure accurate antiderivatives.

This problem requires performing polynomial long division as preparatory algebra before integrating the rational function. The algebraic step is necessary because the degree of the numerator is equal to the degree of the denominator plus one, making the rational function improper. To integrate, we must first divide to express it as a polynomial plus a proper fraction. This allows us to integrate the polynomial terms directly and use logarithmic integration for the fractional remainder. A tempting distractor like choice B fails because it uses a coefficient of 1 instead of 9 for the logarithmic term, resulting from an incorrect remainder in the division. Always perform long division or simplify rational functions before integration to ensure accurate antiderivatives.

This problem requires performing polynomial long division as preparatory algebra before integrating the rational function. The algebraic step is necessary because the degree of the numerator is equal to the degree of the denominator plus one, making the rational function improper. To integrate, we must first divide to express it as a polynomial plus a proper fraction. This allows us to integrate the polynomial terms directly and use logarithmic integration for the fractional remainder. A tempting distractor like choice B fails because it uses a coefficient of 1 instead of 4 for the logarithmic term, resulting from an incorrect remainder in the division. Always perform long division or simplify rational functions before integration to ensure accurate antiderivatives.

This problem requires performing polynomial long division as preparatory algebra before integrating the rational function. The algebraic step is necessary because the degree of the numerator is equal to the degree of the denominator plus one, making the rational function improper. To integrate, we must first divide to express it as a polynomial plus a proper fraction. This allows us to integrate the polynomial terms directly and use logarithmic integration for the fractional remainder. A tempting distractor like choice B fails because it uses a coefficient of 2 instead of 11 for the logarithmic term, resulting from an incorrect remainder in the division. Always perform long division or simplify rational functions before integration to ensure accurate antiderivatives.

This problem requires performing polynomial long division as preparatory algebra before integrating the rational function. The algebraic step is necessary because the degree of the numerator is equal to the degree of the denominator plus one, making the rational function improper. To integrate, we must first divide to express it as a polynomial plus a proper fraction. This allows us to integrate the polynomial terms directly and use logarithmic integration for the fractional remainder. A tempting distractor like choice A fails because it uses a negative coefficient for the logarithmic term, which does not match the positive remainder. Always perform long division or simplify rational functions before integration to ensure accurate antiderivatives.

This problem requires performing polynomial long division as preparatory algebra before integrating the rational function. The algebraic step is necessary because the degree of the numerator is equal to the degree of the denominator plus one, making the rational function improper. To integrate, we must first divide to express it as a polynomial plus a proper fraction. This allows us to integrate the polynomial terms directly and use logarithmic integration for the fractional remainder. A tempting distractor like choice B fails because it uses a coefficient of 1 instead of 4 for the logarithmic term, resulting from an incorrect remainder in the division. Always perform long division or simplify rational functions before integration to ensure accurate antiderivatives.

This problem requires performing polynomial long division as preparatory algebra before integrating the rational function. The algebraic step is necessary because the degree of the numerator is equal to the degree of the denominator plus one, making the rational function improper. To integrate, we must first divide to express it as a polynomial plus a proper fraction. This allows us to integrate the polynomial terms directly and use logarithmic integration for the fractional remainder. A tempting distractor like choice B fails because it uses a coefficient of 16 instead of 20 for the logarithmic term, resulting from an incorrect remainder in the division. Always perform long division or simplify rational functions before integration to ensure accurate antiderivatives.

This problem requires performing polynomial long division as preparatory algebra before integrating the rational function. The algebraic step is necessary because the degree of the numerator is equal to the degree of the denominator plus one, making the rational function improper. To integrate, we must first divide to express it as a polynomial plus a proper fraction. This allows us to integrate the polynomial terms directly and use logarithmic integration for the fractional remainder. A tempting distractor like choice B fails because it uses a coefficient of 1 instead of 4 for the logarithmic term, resulting from an incorrect remainder in the division. Always perform long division or simplify rational functions before integration to ensure accurate antiderivatives.

This problem requires performing polynomial long division as preparatory algebra before integrating the rational function. The algebraic step is necessary because the degree of the numerator is equal to the degree of the denominator plus one, making the rational function improper. To integrate, we must first divide to express it as a polynomial plus a proper fraction. This allows us to integrate the polynomial terms directly and use logarithmic integration for the fractional remainder. A tempting distractor like choice B fails because it uses a coefficient of 10 instead of 8 for the logarithmic term, resulting from an incorrect remainder in the division. Always perform long division or simplify rational functions before integration to ensure accurate antiderivatives.