Precalculus, Chapter 7, 7.4, Section 7.4, Problem 38

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)