Express $\dfrac{x^2+3x+2}{x^2(2x+1)}$ in partial fractions. Hence integrate $\dfrac{x^2+3x+2}{x^2(2x+1)}$ with respect to $x$.

1. Set up the partial fraction decomposition form for a repeated linear factor and a distinct linear factor

Let $\dfrac{x^2+3x+2}{x^2(2x+1)} = \dfrac{A}{x} + \dfrac{B}{x^2} + \dfrac{C}{2x+1}$

2. Multiply by the common denominator to form an identity

$x^2+3x+2 = Ax(2x+1) + B(2x+1) + Cx^2$

3. Substitute $x = 0$ to find $B$

Let $x = 0$:
$0^2 + 3(0) + 2 = A(0) + B(2(0)+1) + C(0)$
$2 = B(1)$
$B = 2$

4. Substitute $x = -\frac{1}{2}$ to find $C$

Let $x = -\frac{1}{2}$:
$(-\frac{1}{2})^2 + 3(-\frac{1}{2}) + 2 = C(-\frac{1}{2})^2$
$\frac{1}{4} - \frac{3}{2} + 2 = \frac{1}{4}C$
$\frac{3}{4} = \frac{1}{4}C$
$C = 3$

5. Compare coefficients of $x^2$ to find $A$

Comparing coefficients of $x^2$:
$1 = 2A + C$
$1 = 2A + 3$
$2A = -2$
$A = -1$

Therefore, $\dfrac{x^2+3x+2}{x^2(2x+1)} = -\dfrac{1}{x} + \dfrac{2}{x^2} + \dfrac{3}{2x+1}$

1. Express the integrand using the partial fractions found in part (a)

From part (a), we have:
$\dfrac{x^2+3x+2}{x^2(2x+1)} = -\dfrac{1}{x} + \dfrac{2}{x^2} + \dfrac{3}{2x+1}$

2. Integrate each term with respect to $x$

$\int \left( -\dfrac{1}{x} + 2x^{-2} + \dfrac{3}{2x+1} \right) dx$
$= -\ln x + \dfrac{2x^{-1}}{-1} + \dfrac{3}{2}\ln(2x+1) + c$
$= -\ln x - \dfrac{2}{x} + \dfrac{3}{2}\ln(2x+1) + c$