`3/(x^4+x)`
Let's factorize the denominator,
`x^4+x=x(x^3+1)`
`=x(x+1)(x^2-x+1)`
Let `3/(x^4+x)=A/x+B/(x+1)+(Cx+D)/(x^2-x+1)`
`3/(x^4+x)=(A(x+1)(x^2-x+1)+B(x)(x^2-x+1)+(Cx+D)(x)(x+1))/(x(x+1)(x^2-x+1))`
`3/(x^4+x)=(A(x^3-x^2+x+x^2-x+1)+B(x^3-x^2+x)+(Cx+D)(x^2+x))/(x(x+1)(x^2-x+1))`
`3/(x^4+x)=(A(x^3+1)+B(x^3-x^2+x)+Cx^3+Cx^2+Dx^2+Dx)/(x(x+1)(x^2-x+1))`
`3/(x^4+x)=(x^3(A+B+C)+x^2(-B+C+D)+x(B+D)+A)/(x(x+1)(x^2-x+1))`
`:.3=x^3(A+B+C)+x^2(-B+C+D)+x(B+D)+A`
equating the coefficients of the like terms,
`A+B+C=0` - equation 1
`-B+C+D=0` - equation 2
`B+D=0` - equation 3
`A=3`
Plug the value of A in equation 1,
`3+B+C=0`
`B+C=-3`
`C=-3-B`
Substitute the above expression of C in equation 2,
`-B+(-3-B)+D=0`
`-B-3-B+D=0`
`-2B+D=3` - equation 4
Now solve equations 3 and...
`3/(x^4+x)`
Let's factorize the denominator,
`x^4+x=x(x^3+1)`
`=x(x+1)(x^2-x+1)`
Let `3/(x^4+x)=A/x+B/(x+1)+(Cx+D)/(x^2-x+1)`
`3/(x^4+x)=(A(x+1)(x^2-x+1)+B(x)(x^2-x+1)+(Cx+D)(x)(x+1))/(x(x+1)(x^2-x+1))`
`3/(x^4+x)=(A(x^3-x^2+x+x^2-x+1)+B(x^3-x^2+x)+(Cx+D)(x^2+x))/(x(x+1)(x^2-x+1))`
`3/(x^4+x)=(A(x^3+1)+B(x^3-x^2+x)+Cx^3+Cx^2+Dx^2+Dx)/(x(x+1)(x^2-x+1))`
`3/(x^4+x)=(x^3(A+B+C)+x^2(-B+C+D)+x(B+D)+A)/(x(x+1)(x^2-x+1))`
`:.3=x^3(A+B+C)+x^2(-B+C+D)+x(B+D)+A`
equating the coefficients of the like terms,
`A+B+C=0` - equation 1
`-B+C+D=0` - equation 2
`B+D=0` - equation 3
`A=3`
Plug the value of A in equation 1,
`3+B+C=0`
`B+C=-3`
`C=-3-B`
Substitute the above expression of C in equation 2,
`-B+(-3-B)+D=0`
`-B-3-B+D=0`
`-2B+D=3` - equation 4
Now solve equations 3 and 4 to get the solutions of B and D,
Subtract equation 3 from equation 4,
`(-2B+D)-(B+D)=3-0`
`-3B=3`
`B=-1`
Plug the value of B in equation 3,
`-1+D=0`
`D=1`
Plug the value of A and B in equation 1,
`3+(-1)+C=0`
`2+C=0`
`C=-2`
`:.3/(x^4+x)=3/x-1/(x+1)+(-2x+1)/(x^2-x+1)`
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