Answer to Question #124539 in Calculus for Ojugbele Daniel

Question #124539
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)
1
Expert's answer
2020-06-30T17:16:10-0400

(i)


3x12x(x2+2)=Ax+Bxx2+2+Cx2+2={3x-1 \over 2x(x^2+2)}={A \over x}+{Bx \over x^2+2}+{C \over x^2+2} =

=Ax2+2A+Bx2+Cx2x(x2+2)={Ax^2+2A+Bx^2+Cx\over 2x(x^2+2)}

A+B=0A+B=0

C=32C=\dfrac{3}{2}

2A=122A=-\dfrac{1}{2}

A=14,B=14,C=32A=-\dfrac{1}{4}, B=\dfrac{1}{4}, C=\dfrac{3}{2}


3x12x(x2+2)=14x+14xx2+2+32x2+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(x1)(x2+1)=Ax1+Bxx2+1+Cx2+1={2x+1 \over (x-1)(x^2+1)}={A \over x-1}+{Bx \over x^2+1}+{C \over x^2+1} =

=Ax2+A+Bx2Bx+CxC(x1)(x2+1)={Ax^2+A+Bx^2-Bx+Cx-C \over (x-1)(x^2+1)}

A+B=0A+B=0

B+C=2-B+C=2

AC=1A-C=1

A=32,B=32,C=12A=\dfrac{3}{2}, B=-\dfrac{3}{2}, C=\dfrac{1}{2}


2x+1(x1)(x2+1)=32x1+32xx2+1+12x2+1{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)


x3+x+1x(x2x+1)=1+x2+1x(x2x+1)={x^3+x+1 \over x(x^2-x+1)}=1+{x^2+1 \over x(x^2-x+1)}=

=1+Ax+Bxx2x+1+Cx2x+1==1+{A \over x}+{Bx \over x^2-x+1}+{C \over x^2-x+1}=

=1+Ax2Ax+A+Bx2+Cxx(x2x+1)=1+{Ax^2-Ax+A+Bx^2+Cx \over x(x^2-x+1)}

A+B=1A+B=1

A+C=0-A+C=0

A=1A=1

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


x3+x+1x(x2x+1)=1+1x+1x2x+1{x^3+x+1 \over x(x^2-x+1)}=1+{1\over x}+{ 1\over x^2-x+1}

(iv)


3x+2(x2)(x2+3x+1){3x+2 \over (x-2)(x^2+3x+1)}


Let x2+3x+1=0x^2+3x+1=0

D=(3)24(1)(1)=5D=(3)^2-4(1)(1)=5

x=3±52x=\dfrac{-3\pm\sqrt{5}}{2}


x2+3x+1=(x352)(x3+52)=x^2+3x+1=(x-\dfrac{-3-\sqrt{5}}{2})(x-\dfrac{-3+\sqrt{5}}{2})=

=14(2x+3+5)(2x+35)={1\over 4}(2x+3+\sqrt{5})(2x+3-\sqrt{5})

3x+2(x2)(x2+3x+1)=Ax2+2B2x+3+5+2C2x+35={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}}=

=4Ax2+12Ax+4A+(x2)(4Bx+6B25B+4Cx+6C+25C)4(x2)(x2+3x+1)={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=04A+4B+4C=0

12A8B+6B25B8C+6C+25C=312A-8B+6B-2\sqrt{5}B-8C+6C+2\sqrt{5}C=3

4A12B+45B12C45C=24A-12B+4\sqrt{5}B-12C-4\sqrt{5}C=2


A+B+C=0A+B+C=0

12B12C2B25B2C+25C=3-12B-12C-2B-2\sqrt{5}B-2C+2\sqrt{5}C=3

16B+45B16C45C=2-16B+4\sqrt{5}B-16C-4\sqrt{5}C=2


A+B+C=0A+B+C=0

14B14C25B+25C=3-14B-14C-2\sqrt{5}B+2\sqrt{5}C=3

8B+25B8C25C=1-8B+2\sqrt{5}B-8C-2\sqrt{5}C=1


A=211A=\dfrac{2}{11}

B=111544B=-\dfrac{1}{11}-\dfrac{\sqrt{5}}{44}

C=111+544C=-\dfrac{1}{11}+\dfrac{\sqrt{5}}{44}


3x+2(x2)(x2+3x+1)=211x2+2115222x+3+5+211+5222x+35{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}}


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