How do you write the partial fraction decomposition of the rational expression $\frac{x^{2} + 1}{x (x + 1) (x^{2} + 2)}$ $(x^2 + 1)/(x(x + 1)(x^2 + 2))$ ?

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How do you write the partial fraction decomposition of the rational expression $\frac{x^{2} + 1}{x (x + 1) (x^{2} + 2)}$ $(x^2 + 1)/(x(x + 1)(x^2 + 2))$ ?

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Answer 1 · 2015-12-14T04:10:04

$\frac{1}{2 x} - \frac{2}{3 (x + 1)} + \frac{x + 2}{6 (x^{2} + 2)}$ $1/(2x)-2/(3(x+1))+(x+2)/(6(x^2+2))$

Explanation:

$\frac{x^{2} + 1}{x (x + 1) (x^{2} + 2)} = \frac{A}{x} + \frac{B}{x + 1} + \frac{C x + D}{x^{2} + 2}$ $(x^2 + 1)/(x(x + 1)(x^2 + 2))=A/x+B/(x+1)+(Cx+D)/(x^2+2)$

$x^{2} + 1 = A (x + 1) (x^{2} + 2) + B (x) (x^{2} + 2) + (C x + D) (x) (x + 1)$ $x^2+1=A(x+1)(x^2+2)+B(x)(x^2+2)+(Cx+D)(x)(x+1)$

$x^{2} + 1 = A x^{3} + A x^{2} + 2 A x + 2 A + B x^{3} + 2 B x + C x^{3} + C x^{2} + D x^{2} + D x$ $x^2+1=Ax^3+Ax^2+2Ax+2A+Bx^3+2Bx+Cx^3+Cx^2+Dx^2+Dx$

$x^{2} + 1 = x^{3} (A + B + C) + x^{2} (A + C + D) + x (2 A + 2 B + D) + 1 (2 A)$ $x^2+1=x^3(A+B+C)+x^2(A+C+D)+x(2A+2B+D)+1(2A)$

Thus, ${(A+B+C=0),(A+C+D=1),(2A+2B+D=0),(2A=1):}$

Solve this to see that ${(A=1/2),(B=-2/3),(C=1/6),(D=1/3):}$

Therefore,

$(x^2 + 1)/(x(x + 1)(x^2 + 2))=1/(2x)-2/(3(x+1))+(x+2)/(6(x^2+2))$

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How do you write the partial fraction decomposition of the rational expression $\frac{x^{2} + 1}{x (x + 1) (x^{2} + 2)}$ $(x^2 + 1)/(x(x + 1)(x^2 + 2))$ ?

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How do you write the partial fraction decomposition of the rational expression (x^2 + 1)/(x(x + 1)(x^2 + 2))x2+1x(x+1)(x2+2)(x^2 + 1)/(x(x + 1)(x^2 + 2))?

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How do you write the partial fraction decomposition of the rational expression $\frac{x^{2} + 1}{x (x + 1) (x^{2} + 2)}$ $(x^2 + 1)/(x(x + 1)(x^2 + 2))$ ?