Seven boys and two girls are to sit together on a bench. In how many ways can they arrange themselves so that the girls do not sit next to each other?

The 7 boys can be arranged in $7!$ ways.

$$7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040$$

When the 7 boys are arranged, there are 8 possible positions (gaps) where the girls can sit so that they are not next to each other.
Let B represent a boy and _ represent a gap:

$$\_ B \_ B \_ B \_ B \_ B \_ B \_ B \_$$

The number of ways to choose 2 distinct gaps out of 8 and arrange the 2 distinct girls in them is given by the permutation formula $P(n, k) = \frac{n!}{(n-k)!}$.

$$P(8, 2) = \frac{8!}{(8-2)!} = \frac{8!}{6!} = 8 \times 7 = 56$$

Multiply the number of ways to arrange the boys by the number of ways to place the girls.

$$\text{Total ways} = 7! \times P(8, 2)$$ $$\text{Total ways} = 5040 \times 56$$ $$\text{Total ways} = 282240$$

The girls can arrange themselves in $\boxed{282240}$ ways so that they do not sit next to each other.

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

The denominator has a linear factor $(3x+1)$ and an irreducible quadratic factor $(1+x^2)$.

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

Multiply both sides by $(3x+1)(1+x^2)$:

$$5x^2 + 2x + 1 = A(1+x^2) + (Bx+C)(3x+1)$$

To find $A$, let $3x+1 = 0$, which means $x = -\frac{1}{3}$. Substitute this value into the equation:

$$5\left(-\frac{1}{3}\right)^2 + 2\left(-\frac{1}{3}\right) + 1 = A\left(1+\left(-\frac{1}{3}\right)^2\right) + \left(B\left(-\frac{1}{3}\right)+C\right)(0)$$ $$5\left(\frac{1}{9}\right) - \frac{2}{3} + 1 = A\left(1+\frac{1}{9}\right)$$ $$\frac{5}{9} - \frac{6}{9} + \frac{9}{9} = A\left(\frac{10}{9}\right)$$ $$\frac{8}{9} = A\left(\frac{10}{9}\right)$$ $$A = \frac{8}{9} \times \frac{9}{10} = \frac{8}{10} = \frac{4}{5}$$

So, $A = \frac{4}{5}$.

Now, expand the right side of the equation and equate coefficients:

$$5x^2 + 2x + 1 = A + Ax^2 + 3Bx^2 + Bx + 3Cx + C$$ $$5x^2 + 2x + 1 = (A+3B)x^2 + (B+3C)x + (A+C)$$

Equating coefficients:
For $x^2$: $A+3B = 5$
For $x$: $B+3C = 2$
Constant: $A+C = 1$

Substitute $A = \frac{4}{5}$ into $A+C=1$:

$$\frac{4}{5} + C = 1$$ $$C = 1 - \frac{4}{5} = \frac{1}{5}$$

So, $C = \frac{1}{5}$.

Substitute $C = \frac{1}{5}$ into $B+3C=2$:

$$B + 3\left(\frac{1}{5}\right) = 2$$ $$B + \frac{3}{5} = 2$$ $$B = 2 - \frac{3}{5} = \frac{10}{5} - \frac{3}{5} = \frac{7}{5}$$

So, $B = \frac{7}{5}$.

Substitute the values of A, B, and C back into the partial fraction form:

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

This can be written as:

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