`(x + 1)/(x^3(x^2 + 1)^2)` Write the partial fraction decomposition of the rational expression. Check your result algebraically.

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You need to decompose the fraction in simple irreducible fractions, such that:

`(x+1)/(x^3(x^2+1)^2)  = A/x + B/(x^2) + C/(x^3) + (Dx+E)/(x^2+1) + (Fx+G)/((x^2+1)^2)`

You need to bring to the same denominator all fractions, such that:

`x+1= A(x^2(x^2+1)^2) + Bx(x^2+1)^2 + C(x^2+1)^2 + (Dx+E)x^3*(x^2+1) + (Fx+G)*x^3`

`x+1= Ax^2(x^4 + 2x^2 + 1) + Bx(x^4 + 2x^2 + 1) + C(x^4 + 2x^2 + 1) + (Dx^4+Ex^3)*(x^2+1) + Fx^4 + Gx^3`

`x+1= Ax^6+ + 2Ax^4 + Ax^2 +Bx^5 + 2Bx^3 + Bx+ Cx^4 + 2Cx^2 +C + Dx^6 + Dx^4 + Ex^5 + Ex^3 + Fx^4 + Gx^3`

You need to group the terms having the same power of x:

`x+1=x^6(A+D) + x^5(B+E) + x^4(2A+C+D+F) + x^3(2B+E+G) + x^2(A+2C) + x(B) + C`

Comparing the expressions both sides yields:

`A+D = 0`

`B+E = 0`

`2A+C+D+F = 0 => F = 1`

`2B+E+G = 0 => G = -1`

`A+2C = 0 => A = -2 => D = 2`

`B= 1 => E = -1`

`C = 1`

Hence, the partial fraction decomposition of the improper rational expression is `(x+1)/(x^3(x^2+1)^2) = -2/x + 1/(x^2) + 1/(x^3) + (2x-1)/(x^2+1) + (x-1)/((x^2+1)^2)`

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