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[edit] Partial fraction

Application of partial fractions : In Integration and In Laplace Transform

Addition and subtraction of ordinary algebraic fractions

Example 1

\frac{2}{x+1}+\frac{3}{x-1}

=\frac{2(x-1)}{(x+1)(x-1)}+\frac{3(x+1)}{(x-1)(x+1)}

=\frac{2(x-1)+3(x+1)}{(x+1)(x-1)}

=\frac{2x-2+3x+3}{(x+1)(x-1)}

=\frac{5x+1}{(x+1)(x-1)}=\frac{Ax+B}{Linear-Denominator}

Exercise 1.1

\frac{2}{x+1}-\frac{3}{x-1}\cdots(1)

\frac{1}{2x+1}+\frac{3}{x-2}\cdots(2)

\frac{3}{2(2x+1)}-\frac{5}{3(x-2)}\cdots(3)

\frac{4}{2x+1}+\frac{5}{x+2}\cdots(4)

Example 2

\frac{2}{x+1}+\frac{3}{(x+1)^2}

=\frac{2(x+1)}{(x+1)^2}+\frac{3}{(x+1)^2}

=\frac{2(x+1)+3}{(x+1)^2}

=\frac{2x+2+3}{(x+1)^2}

=\frac{2x+5}{(x+1)^2}=\frac{Ax+B}{Repeat-Factor-Denominator}

Exercise 1.2

\frac{2}{x+1}-\frac{3}{(x+1)^2}\cdots(1)

\frac{2}{x-1}-\frac{2}{(x-1)^2}\cdots(2)

\frac{3}{x+2}-\frac{5}{(x+2)^2}\cdots(3)

\frac{1}{(2x-1)^2}-\frac{2}{2x-1}\cdots(4)

Example 3

\frac{2}{x+1}+\frac{3}{x^2+1}

=\frac{2(x^2+1)}{(x+1)(x^2+1)}+\frac{3(x+1)}{(x^2+1)(x+1)}

=\frac{2(x^2+1)+3(x+1)}{(x+1)(x^2+1)}

=\frac{2x^2+2+3x+3}{(x+1)(x^2+1)}

=\frac{2x^2+3x+5}{(x+1)(x^2+1)}=\frac{Ax^2+Bx+C}{Quadratic-Denominator}

Exercise 1.3

\frac{2}{x+1}-\frac{3}{x^2+1}\cdots(1)

\frac{2}{x-1}+\frac{3}{x^2+1}\cdots(2)

\frac{2}{x+1}-\frac{2}{x^2-1}\cdots(3)

\frac{5}{2x-1}+\frac{1}{x^2+1}\cdots(4)

Before solving partial fractions. You need to be quick in solving linear and simultaneous equations.

Revision Exercise 1.1

Solve the following simultaneous equations

(1)

3x + y = 4

2x + 3y = 5

(2)

x + y = 2

2x + 3y = 5

(3)

3x - y = 2

2x + 3y = 5

Revision Exercise 1.2

Solve the following linear equations

(1) 3x - 1 = 3

(2) -2x - 1 = -3

(3) -3x + 1 = 4

(4) 2x + 1 = -3

Addition and subtraction of fractions

Revision Exercise 1.3

(1) 2/3 + 1

(2) 1 - 4/6

(3) 1 + 4/6

(4) 5/7 - 3/4

(5) 5/7 + 4/6

Using substitution method and solve the following by finding the constants A , B and C.

(1) x - 1 = A(x + 1)+B(x - 2)

(2) x + 1 = A(x - 1)+B(x + 3)

(3) x + 7 = A(x - 3)+B(x + 4)

(4) x + 3 = A(x - 7)+B(x + 7)

(5) x - 3 = A(x - 4)+B(x + 4)

(6) 2x - 1 = A(x - 6)+B(x + 6)

(7) x² - 3x + 5 = A(x - 2)(x + 3) + B(x + 1)(x + 3) + C(x + 1)(x - 2)

(8) 5x² + 25x - 7 = A(x + 5)(x - 7) + B(x + 3)(x -7 3) + C(x + 3)(x + 5)

Revision Exercise 1.4

Expand the following bracket

(1) (x + 3)(x - 3)

(2) (x - 3)(x - 3)

(3)(2x - 3)(x - 2)

(4) (2x + 3)(2x - 3)

Simplify the following

(1) (x + 3)(x - 3) + (x - 3)(x - 3)

(2) (x + 3)(x - 3) + (2x + 3)(2x - 3)

(3) (2x - 3)(x - 2) + (x + 3)(x - 3)

(4) (2x + 3)(2x - 3) + (2x - 3)(x - 2)

Using equating the coeficient method and solve the following by finding the constants A , B and C.

(1) x - 1 = A(x + 1)+B(x - 2)

(2) x + 1 = A(x - 1)+B(x + 3)

(3) x + 7 = A(x - 3)+B(x + 4)

(4) x + 3 = A(x - 7)+B(x + 7)

(5) x - 3 = A(x - 4)+B(x + 4)

(6) 2x - 1 = A(x - 6)+B(x + 6)

(7) x² - 3x + 5 = A(x - 2)(x + 3) + B(x + 1)(x + 3) + C(x + 1)(x - 2)

(8) 5x² + 25x - 7 = A(x + 5)(x - 7) + B(x + 3)(x -7 3) + C(x + 3)(x + 5)

If substitution method does not work than use the equating the coefficient method