Question

Resolve into partial fraction x^2+1/x^3+1​​

Answer 1

First we have to factorise the denominator
`\rm{x^3+1.}`

It can be factorised as,

`\rm{x^3+1=(x+1)(x^2-x+1)}`

Then the fraction can be written as,

`\longrightarrow\rm{(x^2+1)/(x^3+1)=(x^2+1)/((x+1)(x^2-x+1))\quad\dots(1)}`

This fraction can be resolved into partial fractions by writing it as the sum of a fraction having denominator
`\rm{x+1}` and a fraction having denominator
`\rm{x^2-x+1.}`

Since
`\rm{x+1}` is a first degree polynomial, the numerator of the fraction having this term as denominator should be a zero degree polynomial, or a constant term, assumed as
`\rm{A.}`

Since
`\rm{x^2-x+1}` is a second degree polynomial, the numerator of the fraction having this term as denominator should be a first degree polynomial, assumed as
`\rm{Bx+C.}`

So let,

`\longrightarrow\rm{(x^2+1)/((x+1)(x^2-x+1))=(A)/(x+1)+(Bx+C)/(x^2-x+1)}`

`\small\text{$\longrightarrow\rm{(x^2+1)/((x+1)(x^2-x+1))=(A(x^2-x+1)+(Bx+C)(x+1))/((x+1)(x^2-x+1))}$}`

Equating numerators,

`\longrightarrow\rm{x^2+1=A(x^2-x+1)+(Bx+C)(x+1)}`

`\small\text{$\longrightarrow\rm{x^2+1}=\cal{(A+B)}\rm{x^2}+\cal{(-A+B+C)}\rm{x}+\cal{(A+C)}$}`

Equating corresponding coefficients,

`\rm{A+B=1}`

`\rm{-A+B+C=0}`

`\rm{A+C=1}`

Solving these three equations we get,

`\rm{A=(2)/(3)}`

`\rm{B=(1)/(3)}`

`\rm{C=(1)/(3)}`

Thus,

`\small\text{$\longrightarrow\rm{(x^2+1)/((x+1)(x^2-x+1))=(2)/(3(x+1))+(x+1)/(3(x^2-x+1))}$}`

`\small\text{$\longrightarrow\rm{(x^2+1)/((x+1)(x^2-x+1))=(1)/(3)\left[(2)/(x+1)+(x+1)/(x^2-x+1)\right]}$}`

So (1) becomes,

`\longrightarrow\underline{\underline{\bf{(x^2+1)/(x^3+1)=(1)/(3)\left[(2)/(x+1)+(x+1)/(x^2-x+1)\right]}}}`

Hence the given fraction is resolved into partial fractions.