By using long division, divide $4x^3 + 5x^2 + x - 1$ by $x^2 (x+1)$. Hence, express $\dfrac{4x^3 + 5x^2 + x - 1}{x^2 (x+1)}$ in partial fractions. Hence, find $\displaystyle \int \frac{4x^3 + 5x^2 + x - 1}{x^2 (x+1)}\,dx$.

1. Expand the divisor

Divisor $= x^2(x+1) = x^3 + x^2$

2. Perform long division

$\begin{array}{r|l} x^3 + x^2 & 4x^3 + 5x^2 + x - 1 \\ & 4x^3 + 4x^2 \\ \hline & x^2 + x - 1 \end{array}$

3. Write the result in quotient and remainder form

$\dfrac{4x^3 + 5x^2 + x - 1}{x^3 + x^2} = 4 + \dfrac{x^2 + x - 1}{x^3 + x^2}$

1. Find the partial fraction decomposition from parts (a) and (b)

From part (a), dividing $4x^3 + 5x^2 + x - 1$ by $x^3 + x^2$ gives a quotient of $4$ and a remainder of $x^2 + x - 1$.
So, $\frac{4x^3 + 5x^2 + x - 1}{x^2(x+1)} = 4 + \frac{x^2 + x - 1}{x^2(x+1)}$.
From part (b), let $\frac{x^2 + x - 1}{x^2(x+1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+1}$.
$x^2 + x - 1 = Ax(x+1) + B(x+1) + Cx^2$
Substituting $x = 0$ gives $B = -1$.
Substituting $x = -1$ gives $C = -1$.
Comparing coefficients of $x^2$ gives $1 = A + C \implies A = 2$.
Thus, $\frac{4x^3 + 5x^2 + x - 1}{x^2(x+1)} = 4 + \frac{2}{x} - \frac{1}{x^2} - \frac{1}{x+1}$.

2. Rewrite the integral using the partial fractions

$\displaystyle \int \frac{4x^3 + 5x^2 + x - 1}{x^2 (x+1)}\,dx = \int \left( 4 + \frac{2}{x} - x^{-2} - \frac{1}{x+1} \right)\,dx$

3. Integrate each term to find the final answer

$= 4x + 2\ln x - \frac{x^{-1}}{-1} - \ln(x+1) + c$
$= 4x + 2\ln x + \frac{1}{x} - \ln(x+1) + c$