How do I find the partial-fraction decomposition of $\frac{3 x^{2} + 2 x - 1}{(x + 5) (x^{2} + 1)}$ $(3x^2+2x-1)/((x+5)(x^2+1))$ ?

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How do I find the partial-fraction decomposition of $\frac{3 x^{2} + 2 x - 1}{(x + 5) (x^{2} + 1)}$ $(3x^2+2x-1)/((x+5)(x^2+1))$ ?

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Answer 1 · 2014-10-20T00:19:40

$\frac{3 x^{2} + 2 x - 1}{(x + 5) (x^{2} + 1)}$ $(3x^2+2x-1)/((x+5)(x^2+1))$ becomes ...

$\frac{3 x^{2} + 2 x - 1}{(x + 5) (x^{2} + 1)} = \frac{A}{x + 5} + \frac{B x + C}{x^{2} + 1}$ $(3x^2+2x-1)/((x+5)(x^2+1))=A/(x+5)+(Bx+C)/(x^2+1)$

Multiply through by $(x + 5) (x^{2} + 1)$ $(x+5)(x^2+1)$

$(3 x^{2} + 2 x - 1) = A (x^{2} + 1) + (B x + C) (x + 5)$ $(3x^2+2x-1)=A(x^2+1)+(Bx+C)(x+5)$

First solve for $A$ $A$

$x = - 5$ $x=-5$

$(3 {(- 5)}^{2} + 2 (- 5) - 1) = A ({(- 5)}^{2} + 1) + (B (- 5) + C) ((- 5) + 5)$ $(3(-5)^2+2(-5)-1)=A((-5)^2+1)+(B(-5)+C)((-5)+5)$

$(3 (25) - 10 - 1) = A (25 + 1) + (- 5 B + C) (0)$ $(3(25)-10-1)=A(25+1)+(-5B+C)(0)$

$(75 - 10 - 1) = A (26) + (- 5 B + C) (0)$ $(75-10-1)=A(26)+(-5B+C)(0)$

$64 = 26 A$ $64=26A$

$\frac{64}{26} = \frac{26 A}{26}$ $64/26=(26A)/26$

$\frac{32}{13} = A$ $32/13=A$

Now solve for $C$ $C$

Substitute in for $x$ $x$ and $A$ $A$

$x = 0, A = \frac{32}{13}$ $x=0, A=32/13$

$(3 {(0)}^{2} + 2 (0) - 1) = (\frac{32}{13}) ({(0)}^{2} + 1) + (B (0) + C) ((0) + 5)$ $(3(0)^2+2(0)-1)=(32/13)((0)^2+1)+(B(0)+C)((0)+5)$

$(- 1) = (\frac{32}{13}) (1) + (C) (5)$ $(-1)=(32/13)(1)+(C)(5)$

$- 1 = \frac{32}{13} + 5 C$ $-1=32/13+5C$

$- 1 - \frac{32}{13} = 5 C$ $-1-32/13=5C$

$\frac{- 13}{13} - \frac{32}{13} = 5 C$ $(-13)/13-32/13=5C$

$\frac{- 45}{13} = 5 C$ $(-45)/13=5C$

$\frac{\frac{- 45}{13}}{5} = \frac{5 C}{5}$ $((-45)/13)/5=(5C)/5$

$(\frac{- 45}{13}) \cdot \frac{1}{5} = C$ $((-45)/13)*1/5=C$

$(\frac{- 9}{13}) \cdot \frac{1}{1} = C$ $((-9)/13)*1/1=C$

$\frac{- 9}{13} = C$ $(-9)/13=C$

Substitute in $A$ $A$ and $C$ $C$ and let $x = 1$ $x=1$ as you solve for $B$ $B$

$(3 {(1)}^{2} + 2 (1) - 1) = (\frac{32}{13}) ({(1)}^{2} + 1) + (B (1) + (\frac{- 9}{13})) ((1) + 5)$ $(3(1)^2+2(1)-1)=(32/13)((1)^2+1)+(B(1)+((-9)/13))((1)+5)$

$(3 + 2 - 1) = (\frac{32}{13}) (1 + 1) + (B - \frac{9}{13}) (1 + 5)$ $(3+2-1)=(32/13)(1+1)+(B-9/13)(1+5)$

$4 = (\frac{32}{13}) (2) + (B - \frac{9}{13}) (6)$ $4=(32/13)(2)+(B-9/13)(6)$

$4 = (\frac{64}{13}) + (6 B - \frac{54}{13})$ $4=(64/13)+(6B-54/13)$

$4 - \frac{64}{13} + \frac{54}{13} = 6 B$ $4-64/13+54/13=6B$

$\frac{52}{13} - \frac{64}{13} + \frac{54}{13} = 6 B$ $52/13-64/13+54/13=6B$

$\frac{106}{13} - \frac{64}{13} = 6 B$ $106/13-64/13=6B$

$\frac{42}{13} = 6 B$ $42/13=6B$

$\frac{\frac{42}{13}}{6} = \frac{6 B}{6}$ $(42/13)/6=(6B)/6$

$(\frac{42}{13}) \cdot \frac{1}{6} = B$ $(42/13)*1/6=B$

$\frac{7}{13} = B$ $7/13=B$

Substitute in $A$ $A$ , $B$ $B$ , and $C$ $C$

$\frac{3 x^{2} + 2 x - 1}{(x + 5) (x^{2} + 1)} = \frac{\frac{32}{13}}{x + 5} + \frac{(\frac{7}{13}) x + \frac{- 9}{13}}{x^{2} + 1}$ $(3x^2+2x-1)/((x+5)(x^2+1))=(32/13)/(x+5)+((7/13)x+(-9)/13)/(x^2+1)$

$\frac{3 x^{2} + 2 x - 1}{(x + 5) (x^{2} + 1)} = \frac{32}{13 (x + 5)} + \frac{7 x + (- 9)}{13 (x^{2} + 1)}$ $(3x^2+2x-1)/((x+5)(x^2+1))=32/(13(x+5))+(7x+(-9))/(13(x^2+1))$

Partial-Fraction Decomposition

$\frac{3 x^{2} + 2 x - 1}{(x + 5) (x^{2} + 1)} = \frac{32}{13 (x + 5)} + \frac{7 x - 9}{13 (x^{2} + 1)}$ $(3x^2+2x-1)/((x+5)(x^2+1))=32/(13(x+5))+(7x-9)/(13(x^2+1))$

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How do I find the partial-fraction decomposition of $\frac{3 x^{2} + 2 x - 1}{(x + 5) (x^{2} + 1)}$ $(3x^2+2x-1)/((x+5)(x^2+1))$ ?

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