This integral requires completing the square as the essential preparatory algebra step. The denominator $x^2+4x+8$ must be rewritten in the form $(x-h)^2+k^2$ to enable integration using the arctangent formula. Completing the square: $x^2+4x+8 = (x+2)^2+4 = (x+2)^2+2^2$. This algebraic preparation is crucial because it transforms the integral into the standard form $\int\frac{1}{u^2+a^2}du = \frac{1}{a}\arctan(\frac{u}{a})+C$. With $u=x+2$ and $a=2$, we get $\int\frac{1}{(x+2)^2+4}dx = \frac{1}{2}\arctan(\frac{x+2}{2})+C$. The error of choosing $\ln|x^2+4x+8|+C$ would apply the wrong integration formula for this quadratic form. When the denominator is an irreducible quadratic (discriminant negative), always complete the square to reveal the arctangent structure.

This integral demonstrates the necessity of completing the square as preparatory algebra before integration. We must rewrite $x^2+8x+20$ in standard form to apply the arctangent formula. Completing the square: $x^2+8x+20 = (x+4)^2+4 = (x+4)^2+2^2$. This algebraic preparation transforms the integral into the recognizable form $\int\frac{1}{u^2+a^2}du = \frac{1}{a}\arctan(\frac{u}{a})+C$. With $u=x+4$ and $a=2$, we get $\int\frac{1}{(x+4)^2+4}dx = \frac{1}{2}\arctan(\frac{x+4}{2})+C$. The incorrect choice $\ln|x^2+8x+20|+C$ would result from confusing this with integrals having linear denominators. When integrating rational functions with irreducible quadratic denominators, always complete the square first to expose the underlying arctangent structure.

The key skill here is completing the square as preparatory algebra for integrating rational functions with quadratic denominators. This algebraic manipulation is crucial to transform the denominator into (x - $h)^2$ + k, revealing an arctangent form for the integral. Neglecting it prevents matching the integrand to known antiderivative patterns. The result is (1/3) arctan((x - 2)/3) + C after adjusting the coefficient. A tempting distractor like choice C fails because it lacks the 1/3 scalar, incorrectly scaling the arctangent. Always perform preparatory algebra like completing the square before integrating to ensure the integrand is in a form amenable to basic techniques.

This problem requires algebraic preparation before integration can proceed. The rational function $\frac{x^2+3x+5}{x+1}$ has a numerator of higher degree than the denominator, so polynomial long division must be performed first. Dividing $x^2+3x+5$ by $x+1$ yields quotient $x+2$ with remainder $3$, giving us $\frac{x^2+3x+5}{x+1} = x+2+\frac{3}{x+1}$. Now we can integrate term by term: $\int(x+2+\frac{3}{x+1})dx = \frac{x^2}{2}+2x+3\ln|x+1|+C$. Choice D incorrectly attempts to use logarithm properties without first simplifying the rational expression. The key strategy is to always check if the degree of the numerator is greater than or equal to the denominator's degree—if so, perform polynomial division before integrating.

This integral requires completing the square as the crucial preparatory step. The denominator $x^2-4x+8$ is an irreducible quadratic that must be rewritten in the form $(x-a)^2+b^2$. Completing the square: $x^2-4x+8 = (x-2)^2-4+8 = (x-2)^2+4 = (x-2)^2+2^2$. This transforms our integral to $\int\frac{3}{(x-2)^2+2^2}dx$, which matches the arctangent form $\int\frac{1}{u^2+a^2}du = \frac{1}{a}\arctan(\frac{u}{a})+C$. Applying this with $u=x-2$ and $a=2$ gives $3\cdot\frac{1}{2}\arctan(\frac{x-2}{2})+C = \frac{3}{2}\arctan(\frac{x-2}{2})+C$. Choice C incorrectly assumes this leads to a logarithm rather than arctangent. Always complete the square for irreducible quadratics in the denominator to reveal the appropriate integration form.

This problem demonstrates polynomial long division as the crucial preparatory step for integration. The rational function $\frac{x^2+2x+6}{x+3}$ has numerator degree 2 and denominator degree 1, necessitating division. Performing the division: $x^2+2x+6$ divided by $x+3$ yields quotient $x-1$ with remainder $9$, so $\frac{x^2+2x+6}{x+3} = x-1+\frac{9}{x+3}$. Integrating each term gives $\int(x-1+\frac{9}{x+3})dx = \frac{x^2}{2}-x+9\ln|x+3|+C$. Choice C incorrectly identifies the remainder as 6 instead of 9, likely from a calculation error during polynomial division. The key preparation principle is to systematically perform polynomial division whenever the numerator's degree equals or exceeds the denominator's degree.

This integration problem requires polynomial long division as preparatory algebra since the numerator's degree equals the denominator's degree. Dividing $x^2 - 1$ by $x - 2$ gives quotient $x + 2$ with remainder 3, so $\frac{x^2-1}{x-2} = x + 2 + \frac{3}{x-2}$. This division step is crucial because it transforms the improper rational function into a polynomial plus a simple fraction, each having standard antiderivatives. Integrating term by term yields $\int(x + 2 + \frac{3}{x-2})dx = \frac{1}{2}x^2 + 2x + 3\ln|x-2| + C$. Choice D incorrectly leaves the remainder as $\frac{3}{x-2}$ instead of integrating it to $3\ln|x-2|$, confusing the integrand with its integral. When integrating rational functions, always check if long division is needed by comparing the degrees of numerator and denominator.

This problem requires algebraic preparation through polynomial long division before integrating the rational function. Since the degree of the numerator (2) equals the degree of the denominator (1), we must divide $t^2 + 1$ by $t + 2$ to get $t - 2$ with remainder 5, giving us $\frac{t^2+1}{t+2} = t - 2 + \frac{5}{t+2}$. This decomposition is necessary because we cannot directly integrate the original rational expression—we need it in a form where each piece has a known antiderivative. Integrating term by term yields $\int(t - 2 + \frac{5}{t+2})dt = \frac{1}{2}t^2 - 2t + 5\ln|t+2| + C$. Choice E incorrectly treats the remainder as $\frac{5}{t+2}$ in the final answer rather than integrating it to get $5\ln|t+2|$. Always perform polynomial division when the numerator's degree is greater than or equal to the denominator's degree before attempting integration.

The skill here is using polynomial long division to prepare a rational function for integration. This algebraic manipulation is crucial when the numerator degree is higher than or equal to the denominator's, yielding a polynomial plus a proper rational term. For (2x² + 3x - 1)/(x + 1), division produces 2x + 1 - 2/(x + 1), separating into integrable components. Omitting this step could lead to flawed integration attempts. A tempting distractor is choice B, which replaces the logarithm with -2/(x + 1), but its derivative includes an incorrect squared term in the denominator. Always simplify with long division before integrating rational functions to ensure accuracy.

The skill here is completing the square in the quadratic denominator to prepare for integration. This step is necessary for irreducible quadratics, transforming the expression into a form suitable for the arctangent integral formula. Completing the square on x² + 6x + 13 gives (x + 3)² + 4, allowing a substitution that matches ∫ du/(u² + a²) with a = 2. Without this preparation, one might erroneously apply logarithmic integration rules. A tempting distractor is choice A, which lacks the 1/2 coefficient, failing because its derivative is twice the original integrand. Always complete the square for quadratic denominators before integrating to align with standard forms.