Latihan Gunakan metode pecahan parsial untuk menentukan integral-integral berikut \( 0 \int \frac{x^{3}}{x-5} d x \) \( 0 \int \frac{7 x-7}{x^{2}-5 x-6} d x \) \( \int \frac{3}{(x+1)\left(x^{2}-x\right)} d x \)

We will evaluate the following integrals using the method of partial fractions: 1) ∫ x³⁄(x – 5) dx 2) ∫ (7x – 7)/(x² – 5x – 6) dx 3) ∫ 3⁄[(x + 1)(x² – x)] dx Below are the step‐by‐step solutions. ──────────────────────────── 1) ∫ x³⁄(x – 5) dx Step 1. Since the degree of the numerator (3) is greater than that of the denominator (1), we start with polynomial division. Divide x³ by (x – 5): • Divide: x³ ÷ x = x²   Multiply back: x² · (x – 5) = x³ – 5x² • Subtract:   x³ – (x³ – 5x²) = 5x² • Bring down the “0x” (if you wish to keep the process formal) and continue:   Divide: 5x² ÷ x = 5x   Multiply back: 5x · (x – 5) = 5x² – 25x • Subtract:   5x² – (5x² – 25x) = 25x • Next, divide: 25x ÷ x = 25   Multiply back: 25 · (x – 5) = 25x – 125 • Subtract:   25x – (25x – 125) = 125 Thus, we have:   x³⁄(x – 5) = x² + 5x + 25 + 125⁄(x – 5). Step 2. Now integrate term‐by‐term:   ∫ x³⁄(x – 5) dx = ∫ (x² + 5x + 25) dx + 125 ∫ 1⁄(x – 5) dx Compute the integrals:   ∫ x² dx = (x³)/3   ∫ 5x dx = (5x²)/2   ∫ 25 dx = 25x   ∫ 1⁄(x – 5) dx = ln|x – 5| Thus the answer is:   (x³)/3 + (5x²)/2 + 25x + 125 ln|x – 5| + C ──────────────────────────── 2) ∫ (7x – 7)/(x² – 5x – 6) dx Step 1. Factor the denominator. Notice:   x² – 5x – 6 = (x – 6)(x + 1). Step 2. Write the integrand in partial fractions:   (7x – 7)/[(x – 6)(x + 1)] = A⁄(x – 6) + B⁄(x + 1). Multiply both sides by (x – 6)(x + 1):   7x – 7 = A(x + 1) + B(x – 6). Step 3. Solve for A and B. Expand the right-hand side:   A(x + 1) + B(x – 6) = (A + B)x + (A – 6B). Equate coefficients with 7x – 7:   Coefficient of x: A + B = 7   Constant term: A – 6B = –7 Solve:   From A + B = 7, we get A = 7 – B.   Substitute into A – 6B = –7:    (7 – B) – 6B = –7 ⟹ 7 – 7B = –7 ⟹ –7B = –14 ⟹ B = 2.   Then A = 7 – 2 = 5. Step 4. Write and integrate:   ∫ (7x – 7)/(x² – 5x – 6) dx = 5 ∫ 1⁄(x – 6) dx + 2 ∫ 1⁄(x + 1) dx     = 5 ln|x – 6| + 2 ln|x + 1| + C ──────────────────────────── 3) ∫ 3⁄[(x + 1)(x² – x)] dx Step 1. Factor the quadratic part:   x² – x = x(x – 1). Thus, the denominator becomes:   (x + 1)x(x – 1). Step 2. Write the partial fractions decomposition:   3⁄[x(x – 1)(x + 1)] = A⁄x + B⁄(x – 1) + C⁄(x + 1). Multiply through by x(x – 1)(x + 1):   3 = A (x – 1)(x + 1) + B x(x + 1) + C x(x – 1). Simplify each term:   A (x – 1)(x + 1) = A (x² – 1)   B x(x + 1) = B (x² + x)   C x(x – 1) = C (x² – x). Combine like terms:   3 = (A + B + C)x² + (B – C)x – A. Step 3. Equate coefficients with the constant 3 (which is 0·x² + 0·x + 3):   Coefficient of x²: A + B + C = 0   Coefficient of x:  B – C = 0   Constant term: –A = 3 ⟹ A = –3 From B – C = 0, we have B = C. Then from A + B + C = 0:   –3 + 2B = 0 ⟹ 2B = 3 ⟹ B = 3⁄2, and C = 3⁄2. Step 4. Write the decomposition and integrate:   3⁄[x(x – 1)(x + 1)] = –3⁄x + (3⁄2)⁄(x – 1) + (3⁄2)⁄(x + 1). Thus, the integral becomes:   ∫ 3⁄[(x + 1)(x² – x)] dx = –3 ∫ 1⁄x dx + (3⁄2) ∫ 1⁄(x – 1) dx + (3⁄2) ∫ 1⁄(x + 1) dx     = –3 ln|x| + (3⁄2) ln|x – 1| + (3⁄2) ln|x + 1| + C ──────────────────────────── Final Answers: 1) ∫ x³⁄(x – 5) dx = (x³)/3 + (5x²)/2 + 25x + 125 ln|x – 5| + C 2) ∫ (7x – 7)/(x² – 5x – 6) dx = 5 ln|x – 6| + 2 ln|x + 1| + C 3) ∫ 3⁄[(x + 1)(x² – x)] dx = –3 ln|x| + (3⁄2) ln|x – 1| + (3⁄2) ln|x + 1| + C These are the integrals evaluated using the method of partial fractions.