Evaluate the integral. \[ \int_{1}^{2} \frac{4 y^{2}-2 y-12}{y(y+2)(y-3)} d y \]

We wish to evaluate   I = ∫ from y = 1 to 2 of (4y² – 2y – 12)⁄[y(y + 2)(y – 3)] dy. Step 1. Write the integrand in partial fractions. Assume that   (4y² – 2y – 12)⁄[y(y + 2)(y – 3)] = A/y + B/(y + 2) + C/(y – 3). Multiply both sides by y(y + 2)(y – 3) to clear denominators:   4y² – 2y – 12 = A(y + 2)(y – 3) + B·y(y – 3) + C·y(y + 2). Step 2. Expand and Collect Like Terms. First, expand A(y + 2)(y – 3). Notice that   (y + 2)(y – 3) = y² – y – 6, so  A(y² – y – 6) = A·y² – A·y – 6A. Next, expand B·y(y – 3):   B·y(y – 3) = B·y² – 3B·y. Then, expand C·y(y + 2):   C·y(y + 2) = C·y² + 2C·y. Now, combine all the terms:   A·y² – A·y – 6A + B·y² – 3B·y + C·y² + 2C·y = (A + B + C)y² + (–A – 3B + 2C)y – 6A. Step 3. Equate coefficients with the numerator 4y² – 2y – 12:   Coefficient of y²: A + B + C = 4,   Coefficient of y: –A – 3B + 2C = –2,   Constant term: –6A = –12. From the constant term, solve for A:   –6A = –12 ⟹ A = 2. Now substitute A = 2 into the first equation:   2 + B + C = 4 ⟹ B + C = 2.                       (1) Substitute A = 2 into the y–coefficient equation:   –2 – 3B + 2C = –2 ⟹ –3B + 2C = 0 ⟹ 2C = 3B ⟹ C = (3/2)B.      (2) Now substitute (2) into (1):   B + (3/2)B = 2 ⟹ (5/2)B = 2 ⟹ B = (2·2)/5 = 4/5. Then, from (2), C = (3/2)(4/5) = 6/5. Thus, the partial fractions are:   (4y² – 2y – 12)⁄[y(y + 2)(y – 3)] = 2/y + (4/5)/(y + 2) + (6/5)/(y – 3). Step 4. Integrate Term by Term. The integral becomes:   I = ∫ from 1 to 2 [2/y] dy + (4/5) ∫ from 1 to 2 [1/(y + 2)] dy + (6/5) ∫ from 1 to 2 [1/(y – 3)] dy. Each integral is a standard logarithmic form: • ∫ 1/y dy = ln |y| + C, • ∫ 1/(y + 2) dy = ln |y + 2| + C, • ∫ 1/(y – 3) dy = ln |y – 3| + C. Perform the integrations:   I = 2[ln |y|]₁² + (4/5)[ln |y + 2|]₁² + (6/5)[ln |y – 3|]₁². Evaluate each piece: 1. For 2[ln |y|] from 1 to 2:   = 2 (ln 2 – ln 1) = 2 ln 2   (since ln 1 = 0). 2. For (4/5)[ln |y + 2|] from 1 to 2:   = (4/5) (ln |2 + 2| – ln |1 + 2|) = (4/5)(ln 4 – ln 3). 3. For (6/5)[ln |y – 3|] from 1 to 2:   At y = 2, y – 3 = –1 (absolute value 1) and at y = 1, y – 3 = –2 (absolute value 2). Thus,   = (6/5)(ln 1 – ln 2) = (6/5)(0 – ln 2) = –(6/5) ln 2. Step 5. Combine the results. Add the contributions:   I = 2 ln 2 + (4/5)(ln 4 – ln 3) – (6/5) ln 2. Combine the ln 2 terms:   2 ln 2 – (6/5) ln 2 = (10/5 ln 2 – 6/5 ln 2) = (4/5) ln 2. Thus,   I = (4/5) ln 2 + (4/5)(ln 4 – ln 3)     = (4/5)[ln 2 + ln 4 – ln 3]. Recall that ln 2 + ln 4 = ln (2 · 4) = ln 8. Therefore, the answer is:   I = (4/5)[ln 8 – ln 3] = (4/5) ln(8/3). Final Answer:   ∫ from 1 to 2 (4y² – 2y – 12)⁄[y(y + 2)(y – 3)] dy = (4/5) ln(8/3).