This problem involves using partial fraction decomposition for integration of a rational function with distinct linear factors. To decompose (x+6)/((x+2)(x+5)), write it as A/(x+2) + B/(x+5) and solve for A and B by multiplying through by the denominator, yielding A(x+5) + B(x+2) = x+6. Plugging in x=-2 gives A=4/3, and x=-5 gives B=-1/3. The integral then becomes (4/3) ∫ dx/(x+2) + (-1/3) ∫ dx/(x+5) = (4/3) ln|x+2| - (1/3) ln|x+5| + C. A tempting distractor like choice A ignores the negative sign from the decomposition, failing because it results in a mismatched numerator when verified. In general, for integrating rational functions with distinct linear denominators, decompose into partial fractions, solve the system by plugging in roots or equating coefficients, and integrate each term separately.

The skill used here is partial fraction decomposition to integrate the rational function. Factor the denominator as t² - 4 = (t - 2)(t + 2), and express (5t + 1)/((t - 2)(t + 2)) as A/(t - 2) + B/(t + 2). Solving the equation 5t + 1 = A(t + 2) + B(t - 2) yields A = 11/4 and B = 9/4. Integrating term by term gives (11/4) ln|t - 2| + (9/4) ln|t + 2| + C. A tempting distractor like choice D combines the logs with incorrect coefficients, failing to account for the distinct constants from the decomposition. Always solve for the constants in partial fractions by substituting the roots or equating coefficients to ensure accurate integration of rational functions.

This problem requires the use of partial fraction decomposition to integrate the rational function in an economics model. To decompose 1/((x-2)(x+2)), express it as A/(x-2) + B/(x+2), leading to 1 = A(x+2) + B(x-2). Solving gives A = 1/4 and B = -1/4 by substituting x=2 and x=-2. Integrating each term yields (1/4) ln|x-2| - (1/4) ln|x+2| + C, or equivalently (1/4) ln|(x-2)/(x+2)| + C. A tempting distractor is choice D, which separates with wrong signs, but this fails because it reverses the coefficient signs. In general, for partial fractions with distinct linear factors, solve the system for coefficients accurately and verify by recombining or differentiating the antiderivative.

This problem requires the use of partial fraction decomposition to integrate a rational function. To evaluate ∫ 3 / ((x+1)(x+7)) dx, decompose the integrand as 3 / ((x+1)(x+7)) = A/(x+1) + B/(x+7). Solving for the coefficients gives A = 1/2 and B = -1/2. Integrating term by term yields (1/2) ln |x+1| - (1/2) ln |x+7| + C, which simplifies to (1/2) ln |(x+1)/(x+7)| + C. A tempting distractor is choice B, (1/2) ln |(x+7)/(x+1)| + C, but its derivative is -3 / ((x+1)(x+7)), the negative of the integrand. A transferable partial-fraction strategy is to multiply both sides by the common denominator and solve the resulting linear system for the coefficients.

This problem requires the use of partial fraction decomposition to integrate the rational function in a fluid model. To decompose $\frac{x+4}{x(x+4)}$, express it as $\frac{A}{x} + \frac{B}{x+4}$, leading to $x+4 = A(x+4) + B x$. Solving gives $A = 1$ and $B = 0$ by substituting $x=0$ and $x=-4$, simplifying to $\frac{1}{x}$. Integrating yields $\ln|x| + C$. A tempting distractor is choice A, which includes an extra log, but this fails because the B coefficient is zero, so no second term. In general, for partial fractions with distinct linear factors, solve the system for coefficients accurately and verify by recombining or differentiating the antiderivative.

This problem requires the use of partial fraction decomposition to integrate a rational function. To evaluate $∫ \frac{2}{(x+5)(x-5)} , dx$, decompose the integrand as $\frac{2}{(x+5)(x-5)} = \frac{A}{x+5} + \frac{B}{x-5}$. Solving for the coefficients gives $A = -\frac{1}{5}$ and $B = \frac{1}{5}$. Integrating term by term yields $- \frac{1}{5} \ln |x+5| + \frac{1}{5} \ln |x-5| + C$, which simplifies to $\frac{1}{5} \ln\left| \frac{x-5}{x+5} \right| + C$. A tempting distractor is choice A, $\frac{1}{5} \ln\left| \frac{x+5}{x-5} \right| + C$, but its derivative is $- \frac{2}{(x+5)(x-5)}$, the negative of the integrand. A transferable partial-fraction strategy is to multiply both sides by the common denominator and solve the resulting linear system for the coefficients.

This problem requires the use of partial fraction decomposition to integrate a rational function. To evaluate ∫ (5x+12) / (x(x+6)) dx, decompose the integrand as (5x+12) / (x(x+6)) = A/x + B/(x+6). Solving for the coefficients gives A = 2 and B = 3. Integrating term by term yields 2 ln |x| + 3 ln |x+6| + C. A tempting distractor is choice C, 2 ln |x| - 3 ln |x+6| + C, but its derivative is (5x - 18) / (x(x+6)), not matching 5x+12. A transferable partial-fraction strategy is to multiply both sides by the common denominator and solve the resulting linear system for the coefficients.

This problem requires the use of partial fraction decomposition to integrate the rational function for probability density normalization. To decompose (5x+1)/(x(x+2)), express it as A/x + B/(x+2), leading to 5x+1 = A(x+2) + B x. Solving gives A = 1/2 and B = 9/2 by substituting x=0 and x=-2. Integrating each term yields (1/2) ln|x| + (9/2) ln|x+2| + C. A tempting distractor is choice E, which swaps signs, but this fails because it does not match the positive coefficients from solving. In general, for partial fractions with distinct linear factors, solve the system for coefficients accurately and verify by recombining or differentiating the antiderivative.

This problem uses partial fraction decomposition to integrate a rational function. We write $\frac{2x+1}{(x-1)(x+4)} = \frac{A}{x-1} + \frac{B}{x+4}$ and multiply through by $(x-1)(x+4)$ to get $2x+1 = A(x+4) + B(x-1)$. Setting $x=1$ gives $3 = 5A$, so $A = \frac{3}{5}$; setting $x=-4$ gives $-7 = -5B$, so $B = \frac{7}{5}$. Therefore, $\int \frac{2x+1}{(x-1)(x+4)},dx = \frac{3}{5}\ln|x-1| + \frac{7}{5}\ln|x+4| + C$. Choice E incorrectly assumes equal coefficients of 1 for both logarithms. Always solve for the specific constants A and B rather than guessing.

This integral requires partial fraction decomposition. We write $\frac{6t-1}{(t-3)(t+1)} = \frac{A}{t-3} + \frac{B}{t+1}$ and multiply by $(t-3)(t+1)$ to get $6t-1 = A(t+1) + B(t-3)$. Setting $t=3$ gives $17 = 4A$, so $A = \frac{17}{4}$; setting $t=-1$ yields $-7 = -4B$, so $B = \frac{7}{4}$. Therefore, $\int \frac{6t-1}{(t-3)(t+1)},dt = \frac{17}{4}\ln|t-3| + \frac{7}{4}\ln|t+1| + C$. Choice E incorrectly uses coefficients 6 and -1 without proper decomposition. The key is to clear denominators and evaluate at the roots of each factor.