If there is one prayer that you should pray/sing every day and every hour, it is the LORD's prayer (Our FATHER in Heaven prayer)
It is the most powerful prayer. A pure heart, a clean mind, and a clear conscience is necessary for it. - Samuel Dominic Chukwuemeka

"This is the beginning of the gospel of JESUS CHRIST, the SON of GOD. As it is written in Isaiah the prophet: "Behold, I will send My messenger ahead of You, who will prepare Your way." "A voice of one calling in the wilderness, "Prepare the way for the LORD, make straight paths for Him." John the Baptist appeared in the wilderness, preaching a baptism of repentance for the forgiveness of sins. People went out to him from all Jerusalem and the countryside of Judea. Confessing their sins, they were baptized by him in the Jordan River. John was clothed in camel's hair, with a leather belt around his waist. His food was locusts and wild honey. And he began to proclaim: "After me will come ONE more powerful than I, the straps of whose sandals I am not worthy to stoop down and untie. I baptize you with water, but He will baptize you with the HOLY SPIRIT."" - Mark 1:1 - 8

The Joy of a Teacher is the Success of his Students. - Samuel Chukwuemeka

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 $