Question

Solve the following partial fractions
(i) 3x-1/2x(x^2 +2)

(ii) 2x+1/(x-1)(x^2 +1)

(iii) x^3 +x+1/x(x^2 -x+1)

(iv) 3x+2/(x-2)(x^2 +3x+1)

Answer 1

(i)

$${3x-1 \over 2x(x^2+2)}={A \over x}+{Bx \over x^2+2}+{C \over x^2+2} =$$

$$={Ax^2+2A+Bx^2+Cx\over 2x(x^2+2)}$$

$A+B=0$

$C=\dfrac{3}{2}$

$2A=-\dfrac{1}{2}$

$A=-\dfrac{1}{4}, B=\dfrac{1}{4}, C=\dfrac{3}{2}$

$${3x-1 \over 2x(x^2+2)}={-\dfrac{1}{4} \over x}+{\dfrac{1}{4}x \over x^2+2}+{\dfrac{3}{2}\over x^2+2}$$

(ii)

$${2x+1 \over (x-1)(x^2+1)}={A \over x-1}+{Bx \over x^2+1}+{C \over x^2+1} =$$

$$={Ax^2+A+Bx^2-Bx+Cx-C \over (x-1)(x^2+1)}$$

$A+B=0$

$-B+C=2$

$A-C=1$

$A=\dfrac{3}{2}, B=-\dfrac{3}{2}, C=\dfrac{1}{2}$

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

(iii)

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

$$=1+{A \over x}+{Bx \over x^2-x+1}+{C \over x^2-x+1}=$$

$$=1+{Ax^2-Ax+A+Bx^2+Cx \over x(x^2-x+1)}$$

$A+B=1$

$-A+C=0$

$A=1$

$A=1,B=0,C=1$

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

(iv)

$${3x+2 \over (x-2)(x^2+3x+1)}$$

Let $x^2+3x+1=0$

$D=(3)^2-4(1)(1)=5$

$x=\dfrac{-3\pm\sqrt{5}}{2}$

$$x^2+3x+1=(x-\dfrac{-3-\sqrt{5}}{2})(x-\dfrac{-3+\sqrt{5}}{2})=$$

$$={1\over 4}(2x+3+\sqrt{5})(2x+3-\sqrt{5})$$

$${3x+2 \over (x-2)(x^2+3x+1)}={A \over x-2}+{2B \over 2x+3+\sqrt{5}}+{2C \over 2x+3-\sqrt{5}}=$$

$$={4Ax^2+12Ax+4A+(x-2)(4Bx+6B-2\sqrt{5}B+4Cx+6C+2\sqrt{5}C)\over 4(x-2)(x^2+3x+1)}$$

$4A+4B+4C=0$

$12A-8B+6B-2\sqrt{5}B-8C+6C+2\sqrt{5}C=3$

$4A-12B+4\sqrt{5}B-12C-4\sqrt{5}C=2$

$A+B+C=0$

$-12B-12C-2B-2\sqrt{5}B-2C+2\sqrt{5}C=3$

$-16B+4\sqrt{5}B-16C-4\sqrt{5}C=2$

$A+B+C=0$

$-14B-14C-2\sqrt{5}B+2\sqrt{5}C=3$

$-8B+2\sqrt{5}B-8C-2\sqrt{5}C=1$

$A=\dfrac{2}{11}$

$B=-\dfrac{1}{11}-\dfrac{\sqrt{5}}{44}$

$C=-\dfrac{1}{11}+\dfrac{\sqrt{5}}{44}$

$${3x+2 \over (x-2)(x^2+3x+1)}={\dfrac{2}{11} \over x-2}+{-\dfrac{2}{11}-\dfrac{\sqrt{5}}{22} \over 2x+3+\sqrt{5}}+{-\dfrac{2}{11}+\dfrac{\sqrt{5}}{22} \over 2x+3-\sqrt{5}}$$