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

Oct 20, 2014

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

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

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

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

First solve for $A$

$x = - 5$

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

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

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

$64 = 26 A$

$\frac{64}{26} = \frac{26 A}{26}$

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

Now solve for $C$

Substitute in for $x$ and $A$

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

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

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

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

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

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

$\frac{- 45}{13} = 5 C$

$\frac{\frac{- 45}{13}}{5} = \frac{5 C}{5}$

$\left(\frac{- 45}{13}\right) \cdot \frac{1}{5} = C$

$\left(\frac{- 9}{13}\right) \cdot \frac{1}{1} = C$

$\frac{- 9}{13} = C$

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

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

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

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

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

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

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

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

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

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

$\left(\frac{42}{13}\right) \cdot \frac{1}{6} = B$

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

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

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

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

Partial-Fraction Decomposition

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