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 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 9 instead of 10 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 1 instead of 5 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 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 demonstrates the importance of algebraic simplification before integration. Notice that $x^2-4 = (x-2)(x+2)$, so the fraction $(x^2-4)/(x-2) = (x-2)(x+2)/(x-2) = x+2$ after cancellation. This algebraic preparation transforms a seemingly complex rational function into a simple polynomial that integrates to $\frac{x^2}{2}+2x+C$. Students might mistakenly perform long division and get a logarithmic term (choices A or C), not recognizing the perfect factorization. Always check for factorization opportunities that simplify rational expressions before resorting to long division in integration problems.

This problem requires polynomial long division as essential preparation. Dividing $x^2+2x$ by $x+1$ gives quotient $x+1$ with remainder $-1$, so $(x^2+2x)/(x+1) = x+1-1/(x+1)$. This algebraic step allows us to integrate: $\int(x+1)dx - \int\frac{1}{x+1}dx = \frac{x^2}{2}+x-\ln|x+1|+C$. Without division, you might incorrectly choose option A with a positive logarithm, missing how the remainder creates a subtraction. When integrating rational functions, polynomial long division separates the expression into manageable pieces that follow standard integration formulas.

This problem requires performing polynomial long division or simplification as preparatory algebra before integrating the rational function. The algebraic step is necessary because the rational function can be simplified by factoring and canceling common terms. To integrate, we must first simplify to a polynomial expression. This allows us to integrate the resulting polynomial directly without fractional terms. A tempting distractor like choice A fails because it includes an unnecessary logarithmic term, as if no cancellation occurred. Always perform long division or simplify rational functions before integration to ensure accurate antiderivatives.