Decompose the fraction into partial fractions. $$\frac { 9 x ^ { 2 } + 8 x + 9 } { x ^ { 3 } + x }$$
A) $\quad \frac { 8 } { x } - \frac { 9 } { x ^ { 2 } + 1 }$
B) $\quad \frac { 9 } { x } - \frac { 8 } { x ^ { 2 } + 1 }$
C) $\frac { 8 } { x } + \frac { 9 } { x ^ { 2 } + 1 }$
D) $\frac { 9 } { x } + \frac { 8 } { x ^ { 2 } + 1 }$
E) none of these

Question

Quizplus · Accepted Answer

First, factor the denominator: $x^3 + x = x(x^2 + 1)$. Then, set up the partial fraction decomposition: $\frac{9x^2 + 8x + 9}{x(x^2 + 1)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1}$. Solving for A, B, and C gives A = 9, B = 8, C = 0, leading to $\frac{9}{x} + \frac{8}{x^2 + 1}$.

Decompose the Fraction into Partial Fractions 9x2+8x+9x3+x\frac { 9 x ^ { 2 } + 8 x + 9 } { x ^ { 3 } + x }x3+x9x2+8x+9​

Decompose the Fraction into Partial Fractions $\frac { 9 x ^ { 2 } + 8 x + 9 } { x ^ { 3 } + x }$