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Partial Fractions and Integration — Full Worked Solutions

1. Find ∫ 1/((x+1)(x+3)) dx. Fluency

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   Decompose: 1/((x+1)(x+3)) = A/(x+1) + B/(x+3).

   Multiply both sides by (x+1)(x+3):   1 = A(x+3) + B(x+1).

   x = −1: 1 = 2A ⇒ A = 1/2.

   x = −3: 1 = −2B ⇒ B = −1/2.

   Integrate:

   ∫ 1/((x+1)(x+3)) dx = (1/2)∫ [1/(x+1) − 1/(x+3)] dx

   = (1/2) ln|x+1| − (1/2) ln|x+3| + C

   = (1/2) ln|(x+1)/(x+3)| + C

2. Find ∫ (x+5)/((x−1)(x+2)) dx. Fluency

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   Decompose: (x+5)/((x−1)(x+2)) = A/(x−1) + B/(x+2).

   Multiply: x+5 = A(x+2) + B(x−1).

   x = 1: 6 = 3A ⇒ A = 2.

   x = −2: 3 = −3B ⇒ B = −1.

   Integrate:

   ∫ (x+5)/((x−1)(x+2)) dx = ∫[2/(x−1) − 1/(x+2)] dx

   = 2 ln|x−1| − ln|x+2| + C

   = ln|(x−1)²/(x+2)| + C

3. Find ∫ (3x² + x + 2)/(x(x²+1)) dx. Fluency

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   Decompose: (3x²+x+2)/(x(x²+1)) = A/x + (Bx+C)/(x²+1).

   Multiply by x(x²+1):   3x²+x+2 = A(x²+1) + x(Bx+C).

   Expand: (A+B)x² + Cx + A.

   Constant: A = 2.   Coefficient of x: C = 1.   Coefficient of x²: A+B = 3 ⇒ B = 1.

   Integrate:

   ∫ [2/x + (x+1)/(x²+1)] dx = ∫ 2/x dx + ∫ x/(x²+1) dx + ∫ 1/(x²+1) dx

   = 2 ln|x| + (1/2) ln(x²+1) + arctan(x) + C

   For ∫ x/(x²+1) dx: let u = x²+1, du = 2x dx ⇒ integral = (1/2)ln(x²+1).

   = 2 ln|x| + (1/2) ln(x²+1) + arctan(x) + C

4. Find ∫ 1/(x²−4) dx. Fluency

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   Factorise: x²−4 = (x−2)(x+2).

   Decompose: 1/((x−2)(x+2)) = A/(x−2) + B/(x+2).

   Multiply: 1 = A(x+2) + B(x−2).

   x = 2: 1 = 4A ⇒ A = 1/4.   x = −2: 1 = −4B ⇒ B = −1/4.

   Integrate:

   ∫ 1/(x²−4) dx = (1/4) ln|x−2| − (1/4) ln|x+2| + C

   = (1/4) ln|(x−2)/(x+2)| + C

5. Find ∫ (2x+3)/(x+1)² dx. Understanding

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   Decompose: (2x+3)/(x+1)² = A/(x+1) + B/(x+1)².

   Multiply by (x+1)²: 2x+3 = A(x+1) + B.

   x = −1: 1 = B.   Coefficient of x: 2 = A.

   So A = 2, B = 1.

   Integrate:

   ∫ [2/(x+1) + 1/(x+1)²] dx = 2 ln|x+1| + ∫(x+1)⁻² dx

   = 2 ln|x+1| + (x+1)⁻¹/(−1) + C

   = 2 ln|x+1| − 1/(x+1) + C

6. Investigate ∫₁² (x+1)/(x(x−1)) dx. Understanding

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   Important note: The denominator x(x−1) = 0 at x = 1 (the lower limit), giving a singularity at x = 1.

   Decompose the integrand: (x+1)/(x(x−1)) = A/x + B/(x−1).

   Multiply: x+1 = A(x−1) + Bx.

   x = 0: 1 = −A ⇒ A = −1.   x = 1: 2 = B.

   ∫(x+1)/(x(x−1)) dx = ∫[−1/x + 2/(x−1)] dx = −ln|x| + 2 ln|x−1| + C.

   As x → 1&sup+;: the term 2 ln|x−1| → −∞. The integral diverges.

   Conclusion: The integral ∫₁²(x+1)/(x(x−1)) dx does not converge (improper integral with non-integrable singularity at x = 1).

   This question illustrates the importance of checking for singularities within or at the boundary of the integration interval before evaluating a definite integral.

7. Find ∫ (x²+2)/((x²+1)(x+1)) dx. Understanding

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   Decompose: (x²+2)/((x²+1)(x+1)) = A/(x+1) + (Bx+C)/(x²+1).

   Multiply: x²+2 = A(x²+1) + (Bx+C)(x+1).

   x = −1: 3 = 2A ⇒ A = 3/2.

   Expand: Ax²+A + Bx²+Bx+Cx+C = (A+B)x² + (B+C)x + (A+C).

   x²: A+B = 1 ⇒ B = 1−3/2 = −1/2.   Constant: A+C = 2 ⇒ C = 1/2.

   Check x-coefficient: B+C = −1/2+1/2 = 0 ✓ (numerator has no x-term).

   Integrate:

   ∫ [(3/2)/(x+1) + (−x/2+1/2)/(x²+1)] dx

   = (3/2)ln|x+1| − (1/2)∫ x/(x²+1) dx + (1/2)∫ 1/(x²+1) dx

   = (3/2)ln|x+1| − (1/4)ln(x²+1) + (1/2)arctan(x) + C

   = (3/2)ln|x+1| − (1/4)ln(x²+1) + (1/2)arctan(x) + C

8. Find ∫ (4x²−3x+2)/(x²(x−1)) dx. Understanding

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   Decompose with repeated factor x² = x × x:

   (4x²−3x+2)/(x²(x−1)) = A/x + B/x² + C/(x−1).

   Multiply by x²(x−1): 4x²−3x+2 = Ax(x−1) + B(x−1) + Cx².

   x = 0: 2 = −B ⇒ B = −2.

   x = 1: 4−3+2 = C ⇒ C = 3.

   Coefficient of x²: A+C = 4 ⇒ A = 1.

   Check coefficient of x: −A+B = −1−2 = −3 ✓

   Integrate:

   ∫[1/x − 2/x² + 3/(x−1)] dx

   = ln|x| − 2∫x⁻² dx + 3 ln|x−1|

   = ln|x| + 2/x + 3 ln|x−1| + C

   = ln|x| + 2/x + 3 ln|x−1| + C

9. Find ∫ (x³+1)/(x²−1) dx. Problem Solving

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   Since deg(numerator) = 3 > deg(denominator) = 2, perform polynomial long division.

   Divide x³ + 1 by x² − 1:

   x³ + 1 = x(x²−1) + (x+1). So (x³+1)/(x²−1) = x + (x+1)/(x²−1).

   Factorise denominator: x²−1 = (x−1)(x+1).

   (x+1)/((x−1)(x+1)) = 1/(x−1)   (for x ≠ −1).

   Integrate:

   ∫(x³+1)/(x²−1) dx = ∫x dx + ∫ 1/(x−1) dx

   = x²/2 + ln|x−1| + C

   = x²/2 + ln|x−1| + C

10. Evaluate ∫₀^1/2 1/(1−x²) dx. Problem Solving

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    Factorise: 1−x² = (1−x)(1+x) = −(x−1)(x+1).

    Partial fractions: 1/((1−x)(1+x)) = A/(1−x) + B/(1+x).

    Multiply: 1 = A(1+x) + B(1−x).

    x = 1: 1 = 2A ⇒ A = 1/2.   x = −1: 1 = 2B ⇒ B = 1/2.

    This confirms the identity: 1/(1−x²) = (1/2)[1/(1−x) + 1/(1+x)].

    Evaluate:

    ∫₀^1/2 1/(1−x²) dx = (1/2)[−ln|1−x| + ln|1+x|]₀^1/2

    = (1/2) [ln|(1+x)/(1−x)|]₀^1/2

    At x = 1/2: ln|(3/2)/(1/2)| = ln 3.

    At x = 0: ln|(1)/(1)| = ln 1 = 0.

    = (1/2)(ln 3 − 0)

    = (1/2) ln 3 ≈ 0.549

    Verification using the inverse hyperbolic tangent: ∫ dx/(1−x²) = artanh(x) + C = (1/2)ln((1+x)/(1−x)) + C. At x=1/2: (1/2)ln(3) ✓