`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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