Alg5.1.9

From Exampleproblems

Resolve the expressions into partial fractions $\frac{1-x+6x^2}{x-x^3}\,$

Readjusting the denominator $\frac{1-x+6x^2}{x-x^3}=\frac{1-x+6x^2}{(x)(1-x^2)}\,$

Splitting the denominators into factors, $\frac{1-x+6x^2}{x(1-x^2)}=\frac{1-x+6x^2}{x(1+x)(1-x)}\,$

Let $\frac{1-x+6x^2}{x(1+x)(1-x)}=\frac{A}{x}+\frac{B}{1+x}+\frac{C}{1-x}\,$

Multiplying bothsides by $x(1+x)(1-x)\,$

$(1-x+6x^2)=A(1+x)(1-x)+B(x)(1-x)+C(x)(1+x)\,$

Let $x=1\,$

Substituting the value in the above equation we get $6=2A+2C\,$

Let $6=2A+2C ... (1)\,$

Let $x=0\,$

Substituting the value in the above equation we get $1=A\,$

Let $x=2\,$

Substituting the value in the above equation we get $23=3A+2B+6C\,$

Let $23=3A+2B+6C ... (2)\,$

Substituting the value $A=1\,$ in $(1)\,$

we get $6=2(1)+2C\,$

$2C=4\,$

$C=2\,$

Substituting the values $A=1,C=2\,$ in $(2)\,$

we get $23=3(1)+2B+6(2)\,$

$2B=8\,$

$B=4\,$

Substituting the three values in the main equation we get $\frac{1-x+6x^2}{x-x^3}=\frac{1}{x}+\frac{4}{1+x}+\frac{2}{1-x}\,$