The problem asks us to simplify the expression $\frac{2x^2 + 6x + 1}{x(x+1)}$.

1. Problem Description

The problem asks us to simplify the expression $\frac{2x^2 + 6x + 1}{x(x+1)}$.

2. Solution Steps

We are given the expression $\frac{2x^2 + 6x + 1}{x(x+1)}$.

Let's perform polynomial long division or partial fraction decomposition.

First, let's perform polynomial long division.

However, since the degree of the numerator (2) is equal to the degree of the denominator (after expansion, which is 2), we can rewrite the expression as a sum of simpler fractions.

Assume the following form of partial fraction decomposition:

$$\frac{2x^2 + 6x + 1}{x(x+1)} = A + \frac{B}{x} + \frac{C}{x+1}$$

Multiply both sides by $x(x+1)$:

$$2x^2 + 6x + 1 = A x(x+1) + B(x+1) + Cx$$

$$2x^2 + 6x + 1 = Ax^2 + Ax + Bx + B + Cx$$

$$2x^2 + 6x + 1 = Ax^2 + (A+B+C)x + B$$

Equating the coefficients, we have:

$A = 2$

$A+B+C = 6$

$B = 1$

Substituting $A=2$ and $B=1$ into $A+B+C=6$, we have:

$2 + 1 + C = 6$

$3 + C = 6$

$C = 3$

Therefore,

$$\frac{2x^2 + 6x + 1}{x(x+1)} = 2 + \frac{1}{x} + \frac{3}{x+1}$$

3. Final Answer

$2 + \frac{1}{x} + \frac{3}{x+1}$