Express $\frac{2x^4 + x^3 + 9x^2 + 7x + 3}{(x+1)(x^2+5)}$ in partial fractions.

1. Expand the denominator and perform long division to extract the polynomial part, as the fraction is improper.

Denominator: $(x+1)(x^2+5) = x^3 + x^2 + 5x + 5$

By long division of $(2x^4 + x^3 + 9x^2 + 7x + 3)$ by $(x^3 + x^2 + 5x + 5)$:
$2x(x^3 + x^2 + 5x + 5) = 2x^4 + 2x^3 + 10x^2 + 10x$
Subtracting this from the numerator gives:
$-x^3 - x^2 - 3x + 3$
$-1(x^3 + x^2 + 5x + 5) = -x^3 - x^2 - 5x - 5$
Subtracting this gives the remainder:
$2x + 8$

So, $\frac{2x^4 + x^3 + 9x^2 + 7x + 3}{(x+1)(x^2+5)} = 2x - 1 + \frac{2x + 8}{(x+1)(x^2+5)}$

2. Set up the partial fraction decomposition for the proper fraction remainder.

Let $\frac{2x + 8}{(x+1)(x^2+5)} = \frac{A}{x+1} + \frac{Bx+C}{x^2+5}$

Multiplying by $(x+1)(x^2+5)$:
$2x + 8 = A(x^2+5) + (Bx+C)(x+1)$

3. Solve for the constants $A$, $B$, and $C$.

Substitute $x = -1$:
$2(-1) + 8 = A((-1)^2+5) + 0$
$6 = 6A \implies A = 1$

Substitute $x = 0$:
$8 = 1(5) + C(1)$
$8 = 5 + C \implies C = 3$

Substitute $x = 1$:
$2(1) + 8 = 1(1+5) + (B(1)+3)(1+1)$
$10 = 6 + 2(B+3)$
$4 = 2B + 6 \implies 2B = -2 \implies B = -1$

4. Write the final expression in partial fractions.

$\frac{2x^4 + x^3 + 9x^2 + 7x + 3}{(x+1)(x^2+5)} = 2x - 1 + \frac{1}{x+1} + \frac{3-x}{x^2+5}$