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Solved Examples on Partial Functions

You may verify your answers as applicable with the: Partial Fractions Calculator
Unless specified otherwise:
For each question:
(1.) State the form of the modified whole fraction
(2.) Resolve into partial fractions.
(3.) Show all work.
(4.) Check your solution (if you have time)

(1.) WASSCE:FM

(a.) If $f(x) = \dfrac{2x - 3}{(x^2 - 1)(x + 2)}$

(i) Find all the values of $x$ for which $f(x)$ is undefined.
(ii) Express $f(x)$ in partial fractions.

To find the values of $x$ for which $f(x)$ is undefined, set the denominator to zero and solve for the values of $x$

$ (i) \\[3ex] (x^2 - 1)(x + 2) = 0 \\[3ex] But: \\[3ex] x^2 - 1 = (x + 1)(x - 1)...Difference\;\;of\;\;Two\;\;Squares \\[3ex] \implies \\[3ex] (x + 1)(x - 1)(x + 2) = 0 \\[3ex] x + 1 = 0 \;\;OR\;\; x - 1 = 0 \;\;OR\;\; x + 2 = 0 \\[3ex] x = -1 \;\;OR\;\; x = 1 \;\;OR\;\; x = -2 \\[3ex] $ For the values of $x = -1, 1, 2$; $f(x)$ is undefined

$ (ii) \\[3ex] Numerator:\;\;cannot\;\;be\;\;simplified\;\;further \\[3ex] Denominator = (x^2 - 1)(x + 2) = (x + 1)(x - 1)(x + 2) \\[3ex] \dfrac{2x - 3}{(x^2 - 1)(x + 2)} = \dfrac{2x - 3}{(x + 1)(x - 1)(x + 2)} \\[5ex] $ Form: Proper Fraction: Non-repeated Linear Factors at the Denominator

$ \dfrac{2x - 3}{(x + 1)(x - 1)(x + 2)} \\[5ex] = \dfrac{A}{x + 1} + \dfrac{B}{x - 1} + \dfrac{C}{x + 2} \\[5ex] = \dfrac{A[(x - 1)(x + 2)] + B[(x + 1)(x + 2)] + C[(x + 1)(x - 1)]}{(x + 1)(x - 1)(x + 2)} \\[5ex] = \dfrac{A[x^2 + 2x - x - 2] + B[x^2 + 2x + x + 2] + C[x^2 - 1]}{(x + 1)(x - 1)(x + 2)} \\[5ex] = \dfrac{A(x^2 + x - 2) + B(x^2 + 3x + 2) + Cx^2 - C}{(x + 1)(x - 1)(x + 2)} \\[5ex] = \dfrac{Ax^2 + Ax - 2A + Bx^2 + 3Bx + 2B + Cx^2 - C}{(x + 1)(x - 1)(x + 2)} \\[5ex] = \dfrac{Ax^2 + Bx^2 + Cx^2 + Ax + 3Bx - 2A + 2B - C}{(x + 1)(x - 1)(x + 2)} \\[5ex] $ Denominators are the same
Equate the numerators

$ 2x - 3 = Ax^2 + Bx^2 + Cx^2 + Ax + 3Bx - 2A + 2B - C \\[3ex] 2x - 3 = x^2(A + B + C) + x(A + 3B) - 2A + 2B - C \\[3ex] Swap \\[3ex] x^2(A + B + C) + x(A + 3B) - 2A + 2B - C = 2x - 3 \\[3ex] x^2(A + B + C) + x(A + 3B) - 2A + 2B - C = 0x^2 + 2x - 3 \\[3ex] \implies \\[3ex] A + B + C = 0 ...eqn.(1) \\[3ex] A + 3B = 2 ...eqn.(2) \\[3ex] -2A + 2B - C = -3...eqn.(3) \\[3ex] eqn.(2) - eqn.(1) \implies \\[3ex] (A + 3B) - (A + B + C) = 2 - 0 \\[3ex] A + 3B - A - B - C = 2 \\[3ex] 2B - C = 2 ...eqn.(4) \\[3ex] 2 * eqn.(1) + eqn.(3) \implies \\[3ex] 2(A + B + C) + (-2A + 2B - C) = 2(0) + -3 \\[3ex] 2A + 2B + 2C - 2A + 2B - C = 0 - 3 \\[3ex] 4B + C = -3 ...eqn.(5) \\[3ex] eqn.(4) + eqn.(5) \implies \\[3ex] (2B - C) + (4B + C) = 2 + -3 \\[3ex] 2B - C + 4B + C = 2 - 3 \\[3ex] 6B = -1 \\[3ex] B = \dfrac{-1}{6} \\[5ex] B = -\dfrac{1}{6} \\[5ex] From\;\;eqn.(2) \\[3ex] A + 3B = 2 ...eqn.(2) \\[3ex] A = 2 - 3B \\[3ex] A = 2 - 3\left(-\dfrac{1}{6}\right) \\[5ex] A = 2 + \dfrac{1}{2} \\[5ex] A = \dfrac{4}{2} + \dfrac{1}{2} \\[5ex] A = \dfrac{5}{2} \\[5ex] From\;\;eqn.(4) \\[3ex] 2B - C = 2 ...eqn.(4) \\[3ex] 2B - 2 = C \\[3ex] C = 2B - 2 \\[3ex] C = 2\left(-\dfrac{1}{6}\right) - 2 \\[5ex] C = -\dfrac{1}{3} - \dfrac{2}{1} \\[5ex] C = \dfrac{-1 - 6}{3} \\[5ex] C = -\dfrac{7}{3} \\[5ex] \dfrac{2x - 3}{(x + 1)(x - 1)(x + 2)} = \dfrac{5}{2(x + 1)} + \dfrac{-1}{6(x - 1)} + \dfrac{-7}{3(x + 2)} \\[5ex] \therefore \dfrac{2x - 3}{(x^2 - 1)(x + 2)} = \dfrac{5}{2(x + 1)} - \dfrac{1}{6(x - 1)} - \dfrac{7}{3(x + 2)} \\[5ex] \underline{Check} \\[3ex] \dfrac{2x - 3}{(x^2 - 1)(x + 2)} = \dfrac{5}{2(x + 1)} - \dfrac{1}{6(x - 1)} - \dfrac{7}{3(x + 2)} \\[5ex] \underline{RHS} \\[3ex] \dfrac{5}{2(x + 1)} - \dfrac{1}{6(x - 1)} - \dfrac{7}{3(x + 2)} \\[5ex] = \dfrac{5[3(x - 1)(x + 2)] - 1[(x + 1)(x + 2)] - 7[2(x + 1)(x - 1)]}{6(x + 1)(x - 1)(x + 2)} \\[5ex] = \dfrac{5[3(x^2 + 2x - x - 2)] - 1[(x^2 + 2x + x + 2)] - 7[2(x^2 - 1)]}{6(x + 1)(x - 1)(x + 2)} \\[5ex] = \dfrac{5[3(x^2 + x - 2)] - 1[(x^2 + 3x + 2)] - 7[2x^2 - 2]}{6(x + 1)(x - 1)(x + 2)} \\[5ex] = \dfrac{5[3x^2 + 3x - 6] - x^2 - 3x - 2 - 14x^2 + 14}{6(x + 1)(x - 1)(x + 2)} \\[5ex] = \dfrac{15x^2 + 15x - 30 - x^2 - 3x - 2 - 14x^2 + 14}{6(x + 1)(x - 1)(x + 2)} \\[5ex] = \dfrac{12x - 18}{6(x + 1)(x - 1)(x + 2)} \\[5ex] = \dfrac{6(2x - 3)}{6(x + 1)(x - 1)(x + 2)} \\[5ex] = \dfrac{2x - 3}{(x + 1)(x - 1)(x + 2)} \\[5ex] = \dfrac{2x - 3}{(x^2 - 1)(x + 2)} \\[5ex] = LHS $

(2.) WASSCE:FM

Express $\dfrac{2x - 1}{3x^2 + 4x + 1}$ in partial fractions

$ \underline{Denominator} \\[3ex] 3x^2 + 4x + 1 \\[3ex] = 3x^2 + 3x + x + 1 \\[3ex] = 3x(x + 1) + 1(x + 1) \\[3ex] = (x + 1)(3x + 1) \\[3ex] \implies \\[3ex] \dfrac{2x - 1}{3x^2 + 4x + 1} = \dfrac{2x - 1}{(x + 1)(3x + 1)} \\[5ex] $ Form: Proper Fraction: Non-repeated Linear Factors at the Denominator

$ \dfrac{2x - 1}{(x + 1)(3x + 1)} \\[5ex] = \dfrac{A}{x + 1} + \dfrac{B}{3x + 1} \\[5ex] = \dfrac{A(3x + 1) + B(x + 1)}{(x + 1)(3x + 1)} \\[5ex] = \dfrac{3Ax + A + Bx + B}{(x + 1)(3x + 1)} \\[5ex] = \dfrac{3Ax + Bx + A + B}{(x + 1)(3x + 1)} \\[5ex] $ Denominators are the same
Equate the numerators

$ 2x - 1 = 3Ax + Bx + A + B \\[3ex] 2x - 1 = x(3A + B) + A + B \\[3ex] Swap \\[3ex] x(3A + B) + A + B = 2x - 1 \\[3ex] \implies \\[3ex] 3A + B = 2 ...eqn.(1) \\[3ex] A + B = -1 ...eqn.(2) \\[3ex] eqn.(1) - eqn.(2) \implies \\[3ex] (3A + B) - (A + B) = 2 - (-1) \\[3ex] 3A + B - A - B = 2 + 1 \\[3ex] 2A = 3 \\[3ex] A = \dfrac{3}{2} \\[5ex] From\;\;eqn.(2) \\[3ex] A + B = -1 ...eqn.(2) \\[3ex] B = -1 - A \\[3ex] B = -1 - \dfrac{3}{2} \\[5ex] B = -\dfrac{2}{2} - \dfrac{3}{2} \\[5ex] B = \dfrac{-2 - 3}{2} \\[5ex] B = -\dfrac{5}{2} \\[5ex] \dfrac{2x - 1}{(x + 1)(3x + 1)} = \dfrac{3}{2(x + 1)} + -\dfrac{5}{2(3x + 1)} \\[5ex] \therefore \dfrac{2x - 1}{3x^2 + 4x + 1} = \dfrac{3}{2(x + 1)} - \dfrac{5}{2(3x + 1)} \\[5ex] \underline{Check} \\[3ex] \dfrac{2x - 1}{3x^2 + 4x + 1} = \dfrac{3}{2(x + 1)} - \dfrac{5}{2(3x + 1)} \\[5ex] \underline{RHS} \\[3ex] \dfrac{3}{2(x + 1)} - \dfrac{5}{2(3x + 1)} \\[5ex] = \dfrac{3(3x + 1) - 5(x + 1)}{2(x + 1)(3x + 1)} \\[5ex] = \dfrac{9x + 3 - 5x - 5}{2(x + 1)(3x + 1)} \\[5ex] = \dfrac{4x - 2}{2(x + 1)(3x + 1)} \\[5ex] = \dfrac{2(2x - 1)}{2(x + 1)(3x + 1)} \\[5ex] = \dfrac{2x - 1}{(x + 1)(3x + 1)} \\[5ex] = \dfrac{2x - 1}{3x^2 + 4x + 1} \\[5ex] = LHS $

(3.) WASSCE:FM
(a.) Using the substitution $u = x - 2$, write $\dfrac{x^3 + 5}{(x - 2)^4}$ as an expression in terms of $u$

(b.) Using the answer in $3(a)$, express $\dfrac{x^3 + 5}{(x - 2)^4}$ in partial fractions.

$ (a.) \\[3ex] u = x - 2 \\[3ex] x - 2 = u \\[3ex] x = u + 2 \\[3ex] \implies \\[3ex] \dfrac{x^3 + 5}{(x - 2)^4} = \dfrac{(u + 2)^3 + 5}{u^4} \\[5ex] (b.) \\[3ex] We\;\;have\;\;to\;\;work\;\;with\;\;u...using\;\;the\;\;answer\;\;3(a) \\[3ex] Then,\;\;later\;\;we\;\;express\;\;back\;\;to\;\;x \\[3ex] \dfrac{(u + 2)^3 + 5}{u^4} \\[5ex] (u + 2)^3 = u^3 + 3(u)^2(2) + 3(u)(2)^2 + 2^3...Pascal's\;\;Triangle \\[3ex] (u + 2)^3 = u^3 + 6u^2 + 12u + 8 \\[3ex] Numerator = (u + 2)^3 + 5 \\[3ex] Numerator = u^3 + 6u^2 + 12u + 8 + 5 \\[3ex] Numerator = u^3 + 6u^2 + 12u + 13 \\[3ex] Whole\;\;Fraction = \dfrac{u^3 + 6u^2 + 12u + 13}{u^4} \\[5ex] $ Form: Proper Fraction: Repeated Linear Factors at the Denominator

$ \dfrac{u^3 + 6u^2 + 12u + 13}{u^4} \\[5ex] = \dfrac{A}{u} + \dfrac{B}{u^2} + \dfrac{C}{u^3} + \dfrac{D}{u^4} \\[5ex] = \dfrac{Au^3 + Bu^2 + Cu + D}{u^4} \\[5ex] $ Denominators are the same
Equate the numerators

$ u^3 + 6u^2 + 12u + 13 = Au^3 + Bu^2 + Cu + D \\[3ex] Swap \\[3ex] Au^3 + Bu^2 + Cu + D = u^3 + 6u^2 + 12u + 13 \\[3ex] \implies \\[3ex] A = 1 \\[3ex] B = 6 \\[3ex] C = 12 \\[3ex] D = 13 \\[3ex] \therefore \dfrac{u^3 + 6u^2 + 12u + 13}{u^4} = \dfrac{1}{u} + \dfrac{6}{u^2} + \dfrac{12}{u^3} + \dfrac{13}{u^4} \\[5ex] Express\;\;back\;\;in\;\;x \\[3ex] Recall:\;\; u^3 + 6u^2 + 12u + 13 = x^3 + 5 \\[3ex] Recall:\;\; u = x - 2 \\[3ex] \therefore \dfrac{x^3 + 5}{(x - 2)^4} = \dfrac{1}{x - 2} + \dfrac{6}{(x - 2)^2} + \dfrac{12}{(x - 2)^3} + \dfrac{13}{(x - 2)^4} \\[5ex] \underline{Check} \\[3ex] \dfrac{x^3 + 5}{(x - 2)^4} = \dfrac{1}{x - 2} + \dfrac{6}{(x - 2)^2} + \dfrac{12}{(x - 2)^3} + \dfrac{13}{(x - 2)^4} \\[5ex] \underline{RHS} \\[3ex] \dfrac{1}{x - 2} + \dfrac{6}{(x - 2)^2} + \dfrac{12}{(x - 2)^3} + \dfrac{13}{(x - 2)^4} \\[5ex] = \dfrac{1(x - 2)^3 + 6(x - 2)^2 + 12(x - 2) + 13}{(x - 2)^4} \\[5ex] \underline{Pascal's\;\;Triangle\;\;of\;\;Binomial\;\;Expansion} \\[3ex] (x - 2)^3 = x^3 + 3(x^2)(-2) + 3(x)(-2)^2 + (-2)^3 = x^3 - 6x^2 + 12x - 8 \\[3ex] (x - 2)^2 = x^2 + 2(x)(-2) + (-2)^2 = x^2 - 4x + 4 \\[3ex] Continue \\[3ex] = \dfrac{1(x^3 - 6x^2 + 12x - 8) + 6(x^2 - 4x + 4) + 12x - 24 + 13}{(x - 2)^4} \\[5ex] = \dfrac{x^3 - 6x^2 + 12x - 8 + 6x^2 - 24x + 24 + 12x - 24 + 13}{(x - 2)^4} \\[5ex] = \dfrac{x^3 + 5}{(x - 2)^4} \\[5ex] = LHS $

(4.) WASSCE:FM
A function $f$ is defined by

$f(x) = \dfrac{2x^2 - x + 3}{(x + 1)(x^2 + 1)}$

Express $f(x)$ into partial fractions

Numerator: cannot be simplified further
Denominator: cannot be simplified further
Form: Proper Fraction: Non-repeated Linear and Non-Linear Factors at the Denominator

$ \dfrac{2x^2 - x + 3}{(x + 1)(x^2 + 1)} \\[5ex] = \dfrac{A}{x + 1} + \dfrac{Bx + C}{x^2 + 1} \\[5ex] = \dfrac{A(x^2 + 1) + (Bx + C)(x + 1)}{(x + 1)(x^2 + 1)} \\[5ex] = \dfrac{Ax^2 + A + Bx^2 + Bx + Cx + C}{(x + 1)(x^2 + 1)} \\[5ex] = \dfrac{Ax^2 + Bx^2 + Bx + Cx + A + C}{(x + 1)(x^2 + 1)} \\[5ex] $ Denominators are the same
Equate the numerators

$ 2x^2 - x + 3 = Ax^2 + Bx^2 + Bx + Cx + A + C \\[3ex] 2x^2 - x + 3 = x^2(A + B) + x(B + C) + A + C \\[3ex] Swap \\[3ex] x^2(A + B) + x(B + C) + A + C = 2x^2 - x + 3 \\[3ex] \implies \\[3ex] A + B = 2 ...eqn.(1) \\[3ex] B + C = -1 ...eqn.(2) \\[3ex] A + C = 3 ...eqn.(3) \\[3ex] eqn.(1) - eqn.(2) \implies \\[3ex] (A + B) - (B + C) = 2 - (-1) \\[3ex] A + B - B - C = 2 + 1 \\[3ex] A - C = 3...eqn.(4) \\[3ex] eqn.(3) + eqn.(4) \implies \\[3ex] (A + C) + (A - C) = 3 + 3 \\[3ex] A + C + A - C = 6 \\[3ex] 2A = 6 \\[3ex] A = \dfrac{6}{2} \\[5ex] A = 3 \\[3ex] From\;\;eqn.(1) \\[3ex] A + B = 2 ...eqn.(1) \\[3ex] B = 2 - A \\[3ex] B = 2 - 3 \\[3ex] B = -1 \\[3ex] From\;\;eqn.(3) \\[3ex] A + C = 3 ...eqn.(3) \\[3ex] C = 3 - A \\[3ex] C = 3 - 3 \\[3ex] C = 0 \\[3ex] \dfrac{2x^2 - x + 3}{(x + 1)(x^2 + 1)} = \dfrac{3}{x + 1} + \dfrac{-1x + 0}{x^2 + 1} \\[5ex] \therefore \dfrac{2x^2 - x + 3}{(x + 1)(x^2 + 1)} = \dfrac{3}{x + 1} - \dfrac{x}{x^2 + 1} \\[5ex] \underline{Check} \\[3ex] \dfrac{2x^2 - x + 3}{(x + 1)(x^2 + 1)} = \dfrac{3}{x + 1} - \dfrac{x}{x^2 + 1} \\[5ex] \underline{RHS} \\[3ex] \dfrac{3}{x + 1} - \dfrac{x}{x^2 + 1} \\[5ex] = \dfrac{3(x^2 + 1) - x(x + 1)}{(x + 1)(x^2 + 1)} \\[5ex] = \dfrac{3x^2 + 3 - x^2 - x}{(x + 1)(x^2 + 1)} \\[5ex] = \dfrac{2x^2 - x + 3}{(x + 1)(x^2 + 1)} \\[5ex] = LHS $

(5.) WASSCE:FM

(a) Express $\dfrac{x + 6}{(x + 1)^3}$ in partial fractions

(b) Use the answer in (a) to evaluate $\displaystyle\int_1^2 \dfrac{x + 6}{(x + 1)^3}dx$

$ (a.) \\[3ex] \dfrac{x + 6}{(x + 1)^3} \\[5ex] $ Numerator: cannot be simplified further
Denominator: cannot be simplified further
Form: Proper Fraction: Repeated Linear Factors at the Denominator

$ \dfrac{x + 6}{(x + 1)^3} \\[5ex] = \dfrac{A}{x + 1} + \dfrac{B}{(x + 1)^2} + \dfrac{C}{(x + 1)^3} \\[5ex] = \dfrac{A(x + 1)^2 + B(x + 1) + C}{(x + 1)^3} \\[5ex] = \dfrac{A[(x + 1)(x + 1)] + Bx + B + C}{(x + 1)^3} \\[5ex] = \dfrac{A[x^2 + x + x + 1] + Bx + B + C}{(x + 1)^3} \\[5ex] = \dfrac{A[x^2 + 2x + 1] + Bx + B + C}{(x + 1)^3} \\[5ex] = \dfrac{Ax^2 + 2Ax + A + Bx + B + C}{(x + 1)^3} \\[5ex] = \dfrac{Ax^2 + 2Ax + Bx + A + B + C}{(x + 1)^3} \\[5ex] $ Denominators are the same
Equate the numerators

$ x + 6 = Ax^2 + 2Ax + Bx + A + B + C \\[3ex] Swap \\[3ex] Ax^2 + 2Ax + Bx + A + B + C = x + 6 \\[3ex] Ax^2 + x(2A + B) + A + B + C = 0x^2 + x + 6 \\[3ex] \implies \\[3ex] A = 0...eqn.(1) \\[3ex] 2A + B = 1...eqn.(2) \\[3ex] A + B + C = 6...eqn.(3) \\[3ex] Substitute\;\;eqn.(1)\;\;in\;\;eqn.(2) \\[3ex] 2(0) + B = 1 \\[3ex] 0 + B = 1 \\[3ex] B = 1 - 0 \\[3ex] B = 1 \\[3ex] Substitute\;\;0\;\;for\;\;A\;\;and\;\;1\;\;for\;\;B\;\;in\;\;eqn.(3) \\[3ex] 0 + 1 + C = 6 \\[3ex] 1 + C = 6 \\[3ex] C = 6 - 1 \\[3ex] C = 5 \\[3ex] \implies \\[3ex] \dfrac{x + 6}{(x + 1)^3} = \dfrac{0}{x + 1} + \dfrac{1}{(x + 1)^2} + \dfrac{5}{(x + 1)^3} \\[5ex] \dfrac{x + 6}{(x + 1)^3} = 0 + \dfrac{1}{(x + 1)^2} + \dfrac{5}{(x + 1)^3} \\[5ex] \therefore \dfrac{x + 6}{(x + 1)^3} = \dfrac{1}{(x + 1)^2} + \dfrac{5}{(x + 1)^3} \\[5ex] \underline{Check} \\[3ex] \dfrac{x + 6}{(x + 1)^3} = \dfrac{1}{(x + 1)^2} + \dfrac{5}{(x + 1)^3} \\[5ex] \underline{RHS} \\[3ex] \dfrac{1}{(x + 1)^2} + \dfrac{5}{(x + 1)^3} \\[5ex] = \dfrac{1(x + 1) + 5}{(x + 1)^3} \\[5ex] = \dfrac{x + 1 + 5}{(x + 1)^3} \\[5ex] = \dfrac{x + 6}{(x + 1)^3} \\[5ex] = LHS \\[3ex] (b.) \\[3ex] \displaystyle\int_1^2 \dfrac{x + 6}{(x + 1)^3} dx \\[5ex] = \displaystyle\int_1^2 \left[\dfrac{1}{(x + 1)^2} + \dfrac{5}{(x + 1)^3}\right] dx \\[5ex] = \displaystyle\int_1^2 \dfrac{1}{(x + 1)^2} dx + \displaystyle\int_1^2 \dfrac{5}{(x + 1)^3} dx \\[5ex] Let\;\; p = x + 1 \\[3ex] \dfrac{dp}{dx} = 1 \\[5ex] \dfrac{dx}{dp} = \dfrac{1}{1} = 1 \\[5ex] dx = dp \\[3ex] \displaystyle\int \dfrac{1}{(x + 1)^2} dx \\[5ex] = \displaystyle\int \dfrac{1}{p^2} dp \\[5ex] = \displaystyle\int p^{-2} dp \\[5ex] = \dfrac{p^{-2 + 1}}{-2 + 1} \\[5ex] = \dfrac{p^{-1}}{-1} \\[5ex] = -p^{-1} \\[3ex] = -\dfrac{1}{p} \\[5ex] = -\dfrac{1}{x + 1} \\[5ex] \implies \\[3ex] \displaystyle\int_1^2 \dfrac{1}{(x + 1)^2} dx \\[5ex] = \left[-\dfrac{1}{x + 1}\right]_1^2 \\[5ex] = -\dfrac{1}{2 + 1} - -\dfrac{1}{1 + 1} \\[5ex] = -\dfrac{1}{3} + \dfrac{1}{2} \\[5ex] = \dfrac{-2 + 3}{6} \\[5ex] = \dfrac{1}{6} \\[7ex] \displaystyle\int \dfrac{5}{(x + 1)^3} dx \\[5ex] = 5\displaystyle\int \dfrac{1}{(x + 1)^3} dx \\[5ex] = 5\displaystyle\int \dfrac{1}{p^3} dp \\[5ex] = 5\displaystyle\int p^{-3} dp \\[5ex] = 5 * \dfrac{p^{-3 + 1}}{-3 + 1} \\[5ex] = 5 * \dfrac{p^{-2}}{-2} \\[5ex] = -\dfrac{5}{2} * p^{-2} \\[5ex] = -\dfrac{5}{2} * \dfrac{1}{p^2} \\[5ex] = -\dfrac{5}{2p^2} \\[5ex] = -\dfrac{5}{2(x + 1)^2} \\[5ex] \implies \\[3ex] \displaystyle\int_1^2 \dfrac{5}{(x + 1)^3} dx \\[5ex] = \left[-\dfrac{5}{2(x + 1)^2}\right]_1^2 \\[5ex] = -\dfrac{5}{2(2 + 1)^2} - -\dfrac{5}{2(1 + 1)^2} \\[5ex] = -\dfrac{5}{2(3)^2} + \dfrac{5}{2(2)^2} \\[5ex] = -\dfrac{5}{2(9)} + \dfrac{5}{2(4)} \\[5ex] = -\dfrac{5}{18} + \dfrac{5}{8} \\[5ex] = \dfrac{-20 + 45}{72} \\[5ex] = \dfrac{25}{72} \\[7ex] \implies \\[3ex] \displaystyle\int_1^2 \dfrac{1}{(x + 1)^2} dx + \displaystyle\int_1^2 \dfrac{5}{(x + 1)^3} dx \\[5ex] = \dfrac{1}{6} + \dfrac{25}{72} \\[5ex] = \dfrac{12 + 25}{72} \\[5ex] = \dfrac{37}{72} \\[5ex] \therefore \displaystyle\int_1^2 \dfrac{x + 6}{(x + 1)^3} dx = \dfrac{37}{72} $

(6.) WASSCE:FM
A function $g$ is defined by

$g(x) = \dfrac{3 - 4x}{x^2 + x - 6}$

Express $g(x)$ in partial fractons.

Numerator: cannot be simplified further

$ \underline{Denominator} \\[3ex] x^2 + x - 6 \\[3ex] = (x + 3)(x - 2) \\[3ex] \implies \\[3ex] \dfrac{3 - 4x}{x^2 + x - 6} = \dfrac{3 - 4x}{(x + 3)(x - 2)} \\[5ex] $ Form: Proper Fraction: Non-repeated Linear Factors at the Denominator

$ \dfrac{3 - 4x}{(x + 3)(x - 2)} \\[5ex] = \dfrac{A}{x + 3} + \dfrac{B}{x - 2} \\[5ex] = \dfrac{A(x - 2) + B(x + 3)}{(x + 3)(x - 2)} \\[5ex] = \dfrac{Ax - 2A + Bx + 3B}{(x + 3)(x - 2)} \\[5ex] = \dfrac{Ax + Bx - 2A + 3B}{(x + 3)(x - 2)} \\[5ex] $ Denominators are the same
Equate the numerators

$ 3 - 4x = Ax + Bx - 2A + 3B \\[3ex] -4x + 3 = x(A + B) - 2A + 3B \\[3ex] Swap \\[3ex] x(A + B) - 2A + 3B = -4x + 3 \\[3ex] \implies \\[3ex] A + B = -4 ...eqn.(1) \\[3ex] -2A + 3B = 3 ...eqn.(2) \\[3ex] 2 * eqn.(1) + eqn.(2) \implies \\[3ex] 2(A + B) + (-2A + 3B) = 2(-4) + 3 \\[3ex] 2A + 2B - 2A + 3B = -8 + 3 \\[3ex] 5B = -5 \\[3ex] B = -\dfrac{5}{5} \\[5ex] B = -1 \\[3ex] From\;\;eqn.(1) \\[3ex] A + B = -4 ...eqn.(1) \\[3ex] A = -4 - B \\[3ex] A = -4 - (-1) \\[3ex] A = -4 + 1 \\[3ex] A = -3 \\[3ex] \dfrac{3 - 4x}{(x + 3)(x - 2)} = \dfrac{-3}{x + 3} + -\dfrac{1}{x - 2} \\[5ex] \therefore \dfrac{3 - 4x}{x^2 + x - 6} = -\dfrac{3}{x + 3} - \dfrac{1}{x - 2} \\[5ex] \underline{Check} \\[3ex] \dfrac{3 - 4x}{x^2 + x - 6} = -\dfrac{3}{x + 3} - \dfrac{1}{x - 2} \\[5ex] \underline{RHS} \\[3ex] -\dfrac{3}{x + 3} - \dfrac{1}{x - 2} \\[5ex] = \dfrac{-3(x - 2) - 1(x + 3)}{(x + 3)(x - 2)} \\[5ex] = \dfrac{-3x + 6 - x - 3}{(x + 3)(x - 2)} \\[5ex] = \dfrac{-4x + 3}{(x + 3)(x - 2)} \\[5ex] = \dfrac{3 - 4x}{(x + 3)(x - 2)} \\[5ex] = \dfrac{3 - 4x}{x^2 + x - 6} \\[5ex] = LHS $

(7.) WASSCE:FM

Resolve $\dfrac{x^2 + 1}{(x + 2)^3}$ into partial fractions.

$ \dfrac{x^2 + 1}{(x + 2)^3} \\[5ex] $ Numerator: cannot be simplified further
Denominator: cannot be simplified further
Form: Proper Fraction: Repeated Linear Factors at the Denominator

$ \dfrac{x^2 + 1}{(x + 2)^3} \\[5ex] = \dfrac{A}{x + 2} + \dfrac{B}{(x + 2)^2} + \dfrac{C}{(x + 2)^3} \\[5ex] = \dfrac{A(x + 2)^2 + B(x + 2) + C}{(x + 2)^3} \\[5ex] = \dfrac{A[(x + 2)(x + 2)] + Bx + 2B + C}{(x + 2)^3} \\[5ex] = \dfrac{A[x^2 + 2x + 2x + 4] + Bx + 2B + C}{(x + 2)^3} \\[5ex] = \dfrac{A[x^2 + 4x + 4] + Bx + 2B + C}{(x + 2)^3} \\[5ex] = \dfrac{Ax^2 + 4Ax + 4A + Bx + 2B + C}{(x + 2)^3} \\[5ex] = \dfrac{Ax^2 + 4Ax + Bx + 4A + 2B + C}{(x + 2)^3} \\[5ex] $ Denominators are the same
Equate the numerators

$ x^2 + 1 = Ax^2 + 4Ax + Bx + 4A + 2B + C \\[3ex] Swap \\[3ex] Ax^2 + 4Ax + Bx + 4A + 2B + C = x^2 + 1 \\[3ex] Ax^2 + x(4A + B) + 4A + 2B + C = x^2 + 0x + 1 \\[3ex] \implies \\[3ex] A = 1...eqn.(1) \\[3ex] 4A + B = 0...eqn.(2) \\[3ex] 4A + 2B + C = 1...eqn.(3) \\[3ex] Substitute\;\;eqn.(1)\;\;in\;\;eqn.(2) \\[3ex] 4(1) + B = 0 \\[3ex] 4 + B = 0 \\[3ex] B = 0 - 4 \\[3ex] B = -4 \\[3ex] Substitute\;\;1\;\;for\;\;A\;\;and\;\;-4\;\;for\;\;B\;\;in\;\;eqn.(3) \\[3ex] 4(1) + 2(-4) + C = 1 \\[3ex] 4 - 8 + C = 1 \\[3ex] -4 + C = 1 \\[3ex] C = 1 + 4 \\[3ex] C = 5 \\[3ex] \implies \\[3ex] \dfrac{x^2 + 1}{(x + 2)^3} = \dfrac{1}{x + 2} + \dfrac{-4}{(x + 2)^2} + \dfrac{5}{(x + 2)^3} \\[5ex] \therefore \dfrac{x^2 + 1}{(x + 2)^3} = \dfrac{1}{x + 2} - \dfrac{4}{(x + 2)^2} + \dfrac{5}{(x + 2)^3} \\[5ex] \underline{Check} \\[3ex] \dfrac{x^2 + 1}{(x + 2)^3} = \dfrac{1}{x + 2} - \dfrac{4}{(x + 2)^2} + \dfrac{5}{(x + 2)^3} \\[5ex] \underline{RHS} \\[3ex] \dfrac{1}{x + 2} - \dfrac{4}{(x + 2)^2} + \dfrac{5}{(x + 2)^3} \\[5ex] = \dfrac{1(x + 2)^2 - 4(x + 2) + 5}{(x + 2)^3} \\[5ex] = \dfrac{1[(x + 2)(x + 2)] - 4x - 8 + 5}{(x + 2)^3} \\[5ex] = \dfrac{1(x^2 + 2x + 2x + 4) - 4x - 8 + 5}{(x + 2)^3} \\[5ex] = \dfrac{x^2 + 4x + 4 - 4x - 8 + 5}{(x + 2)^3} \\[5ex] = \dfrac{x^2 + 1}{(x + 2)^3} \\[5ex] = LHS $

(8.) WASSCE:FM

Express $\dfrac{3x + 2}{x^2 + x - 2}$ in partial fractions

Numerator: cannot be simplified further

$ \underline{Denominator} \\[3ex] x^2 + x - 2 \\[3ex] = (x + 2)(x - 1) \\[3ex] \implies \\[3ex] \dfrac{3x + 2}{x^2 + x - 2} = \dfrac{3x + 2}{(x + 2)(x - 1)} \\[5ex] $ Form: Proper Fraction: Non-repeated Linear Factors at the Denominator

$ \dfrac{3x + 2}{(x + 2)(x - 1)} \\[5ex] = \dfrac{A}{x + 2} + \dfrac{B}{x - 1} \\[5ex] = \dfrac{A(x - 1) + B(x + 2)}{(x + 2)(x - 1)} \\[5ex] = \dfrac{Ax - A + Bx + 2B}{(x + 2)(x - 1)} \\[5ex] = \dfrac{Ax + Bx - A + 2B}{(x + 2)(x - 1)} \\[5ex] $ Denominators are the same
Equate the numerators

$ 3x + 2 = Ax + Bx - A + 2B \\[3ex] 3x + 2 = x(A + B) - A + 2B \\[3ex] Swap \\[3ex] x(A + B) - A + 2B = 3x + 2 \\[3ex] \implies \\[3ex] A + B = 3 ...eqn.(1) \\[3ex] -A + 2B = 2 ...eqn.(2) \\[3ex] eqn.(1) + eqn.(2) \implies \\[3ex] 3B = 5 \\[3ex] B = \dfrac{5}{3} \\[5ex] From\;\;eqn.(1) \\[3ex] A + B = 3 ...eqn.(2) \\[3ex] A = 3 - B \\[3ex] A = 3 - \dfrac{5}{3} \\[5ex] A = \dfrac{9}{3} - \dfrac{5}{3} \\[5ex] A = \dfrac{4}{3} \\[5ex] \therefore \dfrac{3x + 2}{x^2 + x - 2} = \dfrac{4}{3(x + 2)} + \dfrac{5}{3(x - 1)} \\[5ex] \underline{Check} \\[3ex] \dfrac{3x + 2}{x^2 + x - 2} = \dfrac{4}{3(x + 2)} + \dfrac{5}{3(x - 1)} \\[5ex] \underline{RHS} \\[3ex] \dfrac{4}{3(x + 2)} + \dfrac{5}{3(x - 1)} \\[5ex] = \dfrac{4(x - 1) + 5(x + 2)}{3(x + 2)(x - 1)} \\[5ex] = \dfrac{4x - 4 + 5x + 10}{3(x + 2)(x - 1)} \\[5ex] = \dfrac{9x + 6}{3(x + 2)(x - 1)} \\[5ex] = \dfrac{3(3x + 2)}{3(x + 2)(x - 1)} \\[5ex] = \dfrac{3x + 2}{(x + 2)(x - 1)} \\[5ex] = \dfrac{3x + 2}{x^2 + x - 2} \\[5ex] = LHS $

(9.) WASSCE:FM

Resolve $\dfrac{2x^2 - 1}{(x^2 - 1)(x - 2)}$ into partial fractions.

$ \dfrac{2x^2 - 1}{(x^2 - 1)(x - 2)} \\[5ex] $ Numerator: "cannot be simplified"
We shall run into radicals if we attempt to use the Difference of Two Squares
Leave as is
Denominator: can be simplified further

$ \underline{Denominator} \\[3ex] x^2 - 1 = (x + 1)(x - 1)...Difference\;\;of\;\;Two\;\;Squares \\[3ex] \implies Denominator = (x + 1)(x - 1)(x - 2) \\[3ex] $ Form: Proper Fraction: Non-repeated Linear Factors at the Denominator

$ \dfrac{2x^2 - 1}{(x + 1)(x - 1)(x - 2)} \\[5ex] = \dfrac{A}{x + 1} + \dfrac{B}{x - 1} + \dfrac{C}{x - 2} \\[5ex] = \dfrac{A[(x - 1)(x - 2)] + B[(x + 1)(x - 2)] + C[(x + 1)(x - 1)]}{(x + 1)(x - 1)(x - 2)} \\[5ex] = \dfrac{A[x^2 - 2x - x + 2] + B[x^2 - 2x + x - 2] + C[x^2 - 1]}{(x + 1)(x - 1)(x - 2)} \\[5ex] = \dfrac{A(x^2 - 3x + 2) + B(x^2 - x - 2) + Cx^2 - C}{(x + 1)(x - 1)(x - 2)} \\[5ex] = \dfrac{Ax^2 - 3Ax + 2A + Bx^2 - Bx - 2B + Cx^2 - C}{(x + 1)(x - 1)(x - 2)} \\[5ex] = \dfrac{Ax^2 + Bx^2 + Cx^2 - 3Ax - Bx + 2A - 2B - C}{(x + 1)(x - 1)(x - 2)} \\[5ex] $ Denominators are the same
Equate the numerators

$ 2x^2 - 1 = Ax^2 + Bx^2 + Cx^2 - 3Ax - Bx + 2A - 2B - C \\[3ex] 2x^2 - 1 = x^2(A + B + C) + x(-3A - B) + 2A - 2B - C \\[3ex] Swap \\[3ex] x^2(A + B + C) + x(-3A - B) + 2A - 2B - C = 2x^2 - 1 \\[3ex] x^2(A + B + C) + x(-3A - B) + 2A - 2B - C = 2x^2 + 0x - 1 \\[3ex] \implies \\[3ex] A + B + C = 2 ...eqn.(1) \\[3ex] -3A - B = 0 ...eqn.(2) \\[3ex] 2A - 2B - C = -1...eqn.(3) \\[3ex] eqn.(1) + eqn.(3) \implies \\[3ex] (A + B + C) + (2A - 2B - C) = 2 + -1 \\[3ex] A + B + C + 2A - 2B - C = 2 - 1 \\[3ex] 3A - B = 1...eqn.(4) \\[3ex] eqn.(2) + eqn.(4) \implies \\[3ex] (-3A - B) + (3A - B) = 0 + 1 \\[3ex] -3A - B + 3A - B = 1 \\[3ex] -2B = 1 \\[3ex] B = -\dfrac{1}{2} \\[5ex] From\;\;eqn.(4) \\[3ex] 3A - B = 1...eqn.(4) \\[3ex] 3A = 1 + B \\[3ex] 3A = 1 + -\dfrac{1}{2} \\[5ex] 3A = 1 - \dfrac{1}{2} \\[5ex] 3A = \dfrac{1}{2} \\[5ex] A = \dfrac{1}{3} * \dfrac{1}{2} \\[5ex] A = \dfrac{1}{6} \\[5ex] From\;\;eqn.(1) \\[3ex] A + B + C = 2 ...eqn.(1) \\[3ex] C = 2 - A - B \\[3ex] C = 2 - \dfrac{1}{6} - -\dfrac{1}{2} \\[5ex] C = \dfrac{2}{1} - \dfrac{1}{6} + \dfrac{1}{2} \\[5ex] C = \dfrac{12 - 1 + 3}{6} \\[5ex] C = \dfrac{14}{6} \\[5ex] C = \dfrac{7}{3} \\[5ex] \dfrac{2x^2 - 1}{(x + 1)(x - 1)(x - 2)} = \dfrac{1}{6(x + 1)} + \dfrac{-1}{2(x - 1)} + \dfrac{7}{3(x - 2)} \\[5ex] \dfrac{2x^2 - 1}{(x + 1)(x - 1)(x - 2)} = \dfrac{1}{6(x + 1)} - \dfrac{1}{2(x - 1)} + \dfrac{7}{3(x - 2)} \\[5ex] \therefore \dfrac{2x^2 - 1}{(x^2 - 1)(x - 2)} = \dfrac{1}{6(x + 1)} - \dfrac{1}{2(x - 1)} + \dfrac{7}{3(x - 2)} \\[5ex] \underline{Check} \\[3ex] \dfrac{2x^2 - 1}{(x^2 - 1)(x - 2)} = \dfrac{1}{6(x + 1)} - \dfrac{1}{2(x - 1)} + \dfrac{7}{3(x - 2)} \\[5ex] \underline{RHS} \\[3ex] \dfrac{1}{6(x + 1)} - \dfrac{1}{2(x - 1)} + \dfrac{7}{3(x - 2)} \\[5ex] = \dfrac{1[(x - 1)(x - 2)] - 1[3(x + 1)(x - 2)] + 7[2(x + 1)(x - 1)]}{6(x + 1)(x - 1)(x - 2)} \\[5ex] = \dfrac{1[x^2 - 2x - x + 2] - 1[3(x^2 - 2x + x - 2)] + 7[2(x^2 - 1)]}{6(x + 1)(x - 1)(x - 2)} \\[5ex] = \dfrac{1[x^2 - 3x + 2] - 1[3(x^2 - x - 2)] + 7[2x^2 - 2]}{6(x + 1)(x - 1)(x - 2)} \\[5ex] = \dfrac{x^2 - 3x + 2 - 1[3x^2 - 3x - 6] + 14x^2 - 14}{6(x + 1)(x - 1)(x - 2)} \\[5ex] = \dfrac{x^2 - 3x + 2 - 3x^2 + 3x + 6 + 14x^2 - 14}{6(x + 1)(x - 1)(x - 2)} \\[5ex] = \dfrac{12x^2 - 6}{6(x + 1)(x - 1)(x - 2)} \\[5ex] = \dfrac{6(2x^2 - 1)}{6(x + 1)(x - 1)(x - 2)} \\[5ex] = \dfrac{2x^2 - 1}{(x + 1)(x - 1)(x - 2)} \\[5ex] = \dfrac{2x^2 - 1}{(x^2 - 1)(x - 2)} \\[5ex] = LHS $

(10.) WASSCE:FM

Express $\dfrac{4x^2 - x + 3}{(x^2 + 1)(x - 1)}$ in partial fractions.

Numerator: cannot be simplified further
Denominator: cannot be simplified further
Form: Proper Fraction: Non-repeated Non-Linear and Linear Factors at the Denominator

$ \dfrac{4x^2 - x + 3}{(x^2 + 1)(x - 1)} \\[5ex] = \dfrac{Ax + B}{x^2 + 1} + \dfrac{C}{x - 1} \\[5ex] = \dfrac{(Ax + B)(x - 1) + C(x^2 + 1)}{(x^2 + 1)(x - 1)} \\[5ex] = \dfrac{Ax^2 - Ax + Bx - B + Cx^2 + C}{(x^2 + 1)(x - 1)} \\[5ex] = \dfrac{Ax^2 + Cx^2 - Ax + Bx - B + C}{(x^2 + 1)(x - 1)} \\[5ex] $ Denominators are the same
Equate the numerators

$ 4x^2 - x + 3 = Ax^2 + Cx^2 - Ax + Bx - B + C \\[3ex] 4x^2 - x + 3 = x^2(A + C) + x(-A + B) - B + C \\[3ex] Swap \\[3ex] x^2(A + C) + x(-A + B) - B + C = 4x^2 - x + 3 \\[3ex] \implies \\[3ex] A + C = 4 ...eqn.(1) \\[3ex] -A + B = -1 ...eqn.(2) \\[3ex] -B + C = 3 ...eqn.(3) \\[3ex] eqn.(1) + eqn.(2) \implies \\[3ex] C + B = 4 + (-1) \\[3ex] B + C = 4 - 1 \\[3ex] B + C = 3...eqn.(4) \\[3ex] eqn.(3) + eqn.(4) \implies \\[3ex] (-B + C) + (B + C) = 3 + 3 \\[3ex] -B + C + B + C = 6 \\[3ex] 2C = 6 \\[3ex] C = \dfrac{6}{2} \\[5ex] C = 3 \\[3ex] From\;\;eqn.(1) \\[3ex] A + C = 4 ...eqn.(1) \\[3ex] A = 4 - C \\[3ex] A = 4 - 3 \\[3ex] A = 1 \\[3ex] From\;\;eqn.(2) \\[3ex] -A + B = -1 ...eqn.(2) \\[3ex] B = -1 + A \\[3ex] B = -1 + 1 \\[3ex] B = 0 \\[3ex] \dfrac{4x^2 - x + 3}{(x^2 + 1)(x - 1)} = \dfrac{1x + 0}{x^2 + 1} + \dfrac{3}{x - 1} \\[5ex] \therefore \dfrac{4x^2 - x + 3}{(x^2 + 1)(x - 1)} = \dfrac{x}{x^2 + 1} + \dfrac{3}{x - 1} \\[5ex] \underline{Check} \\[3ex] \dfrac{4x^2 - x + 3}{(x^2 + 1)(x - 1)} = \dfrac{x}{x^2 + 1} + \dfrac{3}{x - 1} \\[5ex] \underline{RHS} \\[3ex] \dfrac{x}{x^2 + 1} + \dfrac{3}{x - 1} \\[5ex] = \dfrac{x(x - 1) + 3(x^2 + 1)}{(x^2 + 1)(x - 1)} \\[5ex] = \dfrac{x^2 - x + 3x^2 + 3}{(x^2 + 1)(x - 1)} \\[5ex] = \dfrac{4x^2 - x + 3}{(x^2 + 1)(x - 1)} \\[5ex] = LHS $

(11.)

(12.) WASSCE:FM

Express $\dfrac{x + 3}{x^2 - 9x + 18}$ in partial fractions

Numerator: cannot be simplified further

$ \underline{Denominator} \\[3ex] x^2 - 9x + 18 \\[3ex] = (x - 3)(x - 6) \\[3ex] \implies \\[3ex] \dfrac{x + 3}{x^2 - 9x + 18} = \dfrac{x + 3}{(x - 3)(x - 6)} \\[5ex] $ Form: Proper Fraction: Non-repeated Linear Factors at the Denominator

$ \dfrac{x + 3}{(x - 3)(x - 6)} \\[5ex] = \dfrac{A}{x - 3} + \dfrac{B}{x - 6} \\[5ex] = \dfrac{A(x - 6) + B(x - 3)}{(x - 3)(x - 6)} \\[5ex] = \dfrac{Ax - 6A + Bx - 3B}{(x - 3)(x - 6)} \\[5ex] = \dfrac{Ax + Bx - 6A - 3B}{(x - 3)(x - 6)} \\[5ex] $ Denominators are the same
Equate the numerators

$ x + 3 = Ax + Bx - 6A - 3B \\[3ex] x + 3 = x(A + B) - 6A - 3B \\[3ex] Swap \\[3ex] x(A + B) - 6A - 3B = x + 3 \\[3ex] \implies \\[3ex] A + B = 1 ...eqn.(1) \\[3ex] -6A - 3B = 3 ...eqn.(2) \\[3ex] 6 * eqn.(1) + eqn.(2) \implies \\[3ex] 6(A + B) + (-6A - 3B) = 6(1) + 3 \\[3ex] 6A + 6B - 6A - 3B = 6 + 3 \\[3ex] 3B = 9 \\[3ex] B = \dfrac{9}{3} \\[5ex] B = 3 \\[3ex] From\;\;eqn.(1) \\[3ex] A + B = 1 ...eqn.(2) \\[3ex] A = 1 - B \\[3ex] A = 1 - 3 \\[3ex] A = -2 \\[3ex] \dfrac{x + 3}{(x - 3)(x - 6)} = \dfrac{-2}{x - 3} + \dfrac{3}{x - 6} \\[5ex] \therefore \dfrac{x + 3}{x^2 - 9x + 18} = -\dfrac{2}{x - 3} + \dfrac{3}{x - 6} \\[5ex] \underline{Check} \\[3ex] \dfrac{x + 3}{x^2 - 9x + 18} = -\dfrac{2}{x - 3} + \dfrac{3}{x - 6} \\[5ex] \underline{RHS} \\[3ex] -\dfrac{2}{x - 3} + \dfrac{3}{x - 6} \\[5ex] = \dfrac{-2(x - 6) + 3(x - 3)}{(x - 3)(x - 6)} \\[5ex] = \dfrac{-2x + 12 + 3x - 9}{(x - 3)(x - 6)} \\[5ex] = \dfrac{x + 3}{(x - 3)(x - 6)} \\[5ex] = \dfrac{x + 3}{x^2 - 9x + 18} \\[5ex] = LHS $

(13.)

(14.) WASSCE:FM

Express $\dfrac{2x^2 - 5x + 1}{x^3 - 4x^2 + 3x}$ in partial fractions

Numerator: cannot be simplified further

$ \underline{Denominator} \\[3ex] x^3 - 4x^2 + 3x \\[3ex] = x(x^2 - 4x + 3) \\[3ex] = x(x - 1)(x - 3) \\[3ex] \implies \\[3ex] \dfrac{2x^2 - 5x + 1}{x^3 - 4x^2 + 3x} = \dfrac{2x^2 - 5x + 1}{x(x - 1)(x - 3)} \\[5ex] $ Form: Proper Fraction: Non-repeated Linear Factors at the Denominator

$ \dfrac{2x^2 - 5x + 1}{x(x - 1)(x - 3)} \\[5ex] = \dfrac{A}{x} + \dfrac{B}{x - 1} + \dfrac{C}{x - 3} \\[5ex] = \dfrac{A[(x - 1)(x - 3)] + Bx(x - 3) + Cx(x - 1)}{x(x - 1)(x - 3)} \\[5ex] = \dfrac{A[x^2 - 3x - x + 3] + Bx^2 - 3Bx + Cx^2 - Cx}{x(x - 1)(x - 3)} \\[5ex] = \dfrac{A(x^2 - 4x + 3) + Bx^2 + Cx^2 - 3Bx - Cx}{x(x - 1)(x - 3)} \\[5ex] = \dfrac{Ax^2 - 4Ax + 3A + Bx^2 + Cx^2 - 3Bx - Cx}{x(x - 1)(x - 3)} \\[5ex] = \dfrac{Ax^2 + Bx^2 + Cx^2 - 4Ax - 3Bx - Cx + 3A}{x(x - 1)(x - 3)} \\[5ex] $ Denominators are the same
Equate the numerators

$ 2x^2 - 5x + 1 = Ax^2 + Bx^2 + Cx^2 - 4Ax - 3Bx - Cx + 3A \\[3ex] 2x^2 - 5x + 1 = x^2(A + B + C) + x(-4A - 3B - C) + 3A \\[3ex] Swap \\[3ex] x^2(A + B + C) + x(-4A - 3B - C) + 3A = 2x^2 - 5x + 1 \\[3ex] \implies \\[3ex] A + B + C = 2 ...eqn.(1) \\[3ex] -4A - 3B - C = -5 ...eqn.(2) \\[3ex] eqn.(2) \implies -(4A + 3B + C) = -5 \\[3ex] eqn.(2) \implies 4A + 3B + C = 5 \\[3ex] 4A + 3B + C = 5 ...eqn.(2) \\[3ex] 3A = 1...eqn.(3) \\[3ex] From\;\;eqn.(3) \\[3ex] A = \dfrac{1}{3} \\[5ex] eqn.(2) - eqn.(1) \implies \\[3ex] (4A + 3B + C) - (A + B + C) = 5 - 2 \\[3ex] 4A + 3B + C - A - B - C = 3 \\[3ex] 3A + 2B = 3...eqn.(4) \\[3ex] Substitute\;\; \dfrac{1}{3}\;\;for\;\;A\;\;in\;\;eqn.(4) \\[5ex] 3\left(\dfrac{1}{3}\right) + 2B = 3 \\[3ex] 1 + 2B = 3 \\[3ex] 2B = 3 - 1 \\[3ex] 2B = 2 \\[3ex] B = \dfrac{2}{2} \\[5ex] B = 1 \\[3ex] Substitute\;\; \dfrac{1}{3}\;\;for\;\;A\;\;and\;\;1\;\;for\;\;B\;\;in\;\;eqn.(1) \\[5ex] From\;\;eqn.(1) \\[3ex] A + B + C = 2 ...eqn.(1) \\[3ex] C = 2 - A - B \\[3ex] C = 2 - \dfrac{1}{3} - 1 \\[3ex] C = \dfrac{6}{3} - \dfrac{1}{3} - \dfrac{3}{3} \\[5ex] C = \dfrac{2}{3} \\[5ex] \dfrac{2x^2 - 5x + 1}{x(x - 1)(x - 3)} = \dfrac{1}{3x} + \dfrac{1}{x - 1} + \dfrac{2}{3(x - 3)} \\[5ex] \therefore \dfrac{2x^2 - 5x + 1}{x^3 - 4x^2 + 3x} = \dfrac{1}{3x} + \dfrac{1}{x - 1} + \dfrac{2}{3(x - 3)} \\[5ex] \underline{Check} \\[3ex] \dfrac{2x^2 - 5x + 1}{x^3 - 4x^2 + 3x} = \dfrac{1}{3x} + \dfrac{1}{x - 1} + \dfrac{2}{3(x - 3)} \\[5ex] \underline{RHS} \\[3ex] \dfrac{1}{3x} + \dfrac{1}{x - 1} + \dfrac{2}{3(x - 3)} \\[5ex] = \dfrac{1[(x - 1)(x - 3)] + 1[3x(x - 3)] + 2x(x - 1)}{3x(x - 1)(x - 3)} \\[5ex] = \dfrac{(x^2 - 3x - x + 3) + 3x^2 - 9x + 2x^2 - 2x}{3x(x - 1)(x - 3)} \\[5ex] = \dfrac{x^2 - 4x + 3 + 3x^2 - 9x + 2x^2 - 2x}{3x(x - 1)(x - 3)} \\[5ex] = \dfrac{6x^2 - 15x + 3}{3x(x - 1)(x - 3)} \\[5ex] = \dfrac{3(2x^2 - 5x + 1)}{3x(x - 1)(x - 3)} \\[5ex] = \dfrac{2x^2 - 5x + 1}{x(x - 1)(x - 3)} \\[5ex] = \dfrac{2x^2 - 5x + 1}{x^3 - 4x^2 + 3x} \\[5ex] = LHS $

(16.) ATAR
(a) Given that $\dfrac{2}{(x + 1)(x - 1)} = \dfrac{a}{x - 1} + \dfrac{b}{x + 1}$

determine the values for a and b

(b) Hence determine $\displaystyle\int \dfrac{1}{x^2 - 1}dx$

Numerator: cannot be simplified further

Form: Proper Fraction: Non-repeated Linear Factors at the Denominator

$ \dfrac{2}{(x + 1)(x - 1)} \\[5ex] = \dfrac{a}{x - 1} + \dfrac{b}{x + 1} \\[5ex] = \dfrac{a(x + 1) + b(x - 1)}{(x - 1)(x + 1)} \\[5ex] = \dfrac{ax + a + bx - b}{(x - 1)(x + 1)} \\[5ex] = \dfrac{ax + bx + a - b}{(x - 1)(x + 1)} \\[5ex] $ Denominators are the same
Equate the numerators

$ 2 = ax + bx + a - b \\[3ex] ax + bx + a - b = 2 \\[3ex] ax + bx + a - b = 0x + 2 \\[3ex] \implies \\[3ex] a + b = 0 ...eqn.(1) \\[3ex] a - b = 2 ...eqn.(2) \\[3ex] eqn.(1) + eqn.(2) \implies \\[3ex] 2a = 2 \\[3ex] a = \dfrac{2}{2} \\[5ex] a = 1 \\[3ex] eqn.(1) - eqn.(2) \implies \\[3ex] 2b = -2 \\[3ex] b = -\dfrac{2}{2} \\[5ex] b = -1 \\[3ex] \therefore \dfrac{2}{(x + 1)(x - 1)} = \dfrac{1}{x - 1} + \dfrac{-1}{x + 1} \\[5ex] \dfrac{2}{(x - 1)(x + 1)} = \dfrac{1}{x - 1} - \dfrac{1}{x + 1} \\[5ex] \underline{Check} \\[3ex] \underline{RHS} \\[3ex] \dfrac{1}{x - 1} - \dfrac{1}{x + 1} \\[5ex] = \dfrac{1(x + 1) - 1(x - 1)}{(x - 1)(x + 1)} \\[5ex] = \dfrac{x + 1 - x + 1}{(x - 1)(x + 1)} \\[5ex] = \dfrac{2}{(x - 1)(x + 1)} \\[5ex] = \dfrac{2}{(x + 1)(x - 1)} \\[5ex] = LHS \\[3ex] $ (b) Because of the word, "Hence"; we are expected to use the result we got from Part (a) to solve Part (b)
However, the partial fraction we did in (a) is slightly different from the integral in (b)
So, we need to make some modifications in order to solve it.

$ \displaystyle\int \dfrac{1}{x^2 - 1}dx \\[5ex] \underline{Denominator} \\[3ex] x^2 - 1 = (x + 1)(x - 1) ...Difference\;\;of\;\;Two\;\;Squares \\[3ex] \displaystyle\int \dfrac{1}{x^2 - 1}dx \\[5ex] = \displaystyle\int \dfrac{1}{(x + 1)(x - 1)} dx \\[5ex] = \dfrac{1}{2} * \displaystyle\int \dfrac{2}{(x + 1)(x - 1)} dx \\[5ex] = \dfrac{1}{2} * \left[\displaystyle\int \left(\dfrac{1}{x - 1} - \dfrac{1}{x + 1}\right) dx\right] \\[5ex] = \dfrac{1}{2} * \left[\displaystyle\int \dfrac{1}{x - 1}dx - \displaystyle\int \dfrac{1}{x + 1}dx\right] \\[5ex] = \dfrac{1}{2}[\ln|x - 1| - \ln|x + 1|] + C \\[5ex] = \dfrac{1}{2}\ln\dfrac{|x - 1|}{|x + 1|} + C $