Express $\frac{2x^3+2x^2-7x+4}{x(x-1)^2}$ in partial fractions. Hence evaluate $\int_{2}^{4} \frac{4x^3+4x^2-14x+8}{3x(x-1)^2} \, dx$.

1. Recognise that the fraction is improper and perform long division to extract the polynomial part.

The degree of the numerator is 3 and the degree of the denominator is $x(x-1)^2 = x^3 - 2x^2 + x$, which is also 3. Thus, the fraction is improper.

By long division or inspection:
$2x^3 + 2x^2 - 7x + 4 = 2(x^3 - 2x^2 + x) + (6x^2 - 9x + 4)$

So, $\frac{2x^3+2x^2-7x+4}{x(x-1)^2} = 2 + \frac{6x^2-9x+4}{x(x-1)^2}$

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

Let $\frac{6x^2-9x+4}{x(x-1)^2} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{(x-1)^2}$

Multiplying throughout by $x(x-1)^2$ gives:
$6x^2 - 9x + 4 = A(x-1)^2 + Bx(x-1) + Cx$

3. Substitute suitable values of $x$ to find constants $A$ and $C$.

Substitute $x = 0$:
$4 = A(-1)^2 \implies A = 4$

Substitute $x = 1$:
$6(1)^2 - 9(1) + 4 = C(1) \implies C = 1$

4. Compare coefficients to find the remaining constant $B$ and state the final answer.

Compare coefficients of $x^2$:
$6 = A + B$
$6 = 4 + B \implies B = 2$

Therefore, the partial fractions are:
$2 + \frac{4}{x} + \frac{2}{x-1} + \frac{1}{(x-1)^2}$

1. Relate the integrand to the expression in part (a)

Notice that the integrand is a multiple of the expression in part (a):
$\frac{4x^3+4x^2-14x+8}{3x(x-1)^2} = \frac{2}{3} \left( \frac{2x^3+2x^2-7x+4}{x(x-1)^2} \right)$

2. Substitute the partial fractions from part (a)

From part (a), we have $\frac{2x^3+2x^2-7x+4}{x(x-1)^2} = 2 + \frac{4}{x} + \frac{2}{x-1} + \frac{1}{(x-1)^2}$.
So the integral becomes:
$\frac{2}{3} \int_{2}^{4} \left( 2 + \frac{4}{x} + \frac{2}{x-1} + (x-1)^{-2} \right) \, dx$

3. Integrate the expression with respect to $x$

$\frac{2}{3} \left[ 2x + 4\ln|x| + 2\ln|x-1| - \frac{1}{x-1} \right]_{2}^{4}$

4. Evaluate the definite integral by substituting the limits

Substitute $x = 4$:
$2(4) + 4\ln 4 + 2\ln 3 - \frac{1}{3} = \frac{23}{3} + 8\ln 2 + 2\ln 3$

Substitute $x = 2$:
$2(2) + 4\ln 2 + 2\ln 1 - 1 = 3 + 4\ln 2$

Subtracting the lower limit value from the upper limit value:
$\frac{2}{3} \left( \left( \frac{23}{3} + 8\ln 2 + 2\ln 3 \right) - (3 + 4\ln 2) \right)$
$= \frac{2}{3} \left( \frac{14}{3} + 4\ln 2 + 2\ln 3 \right)$
$= \frac{28}{9} + \frac{8}{3}\ln 2 + \frac{4}{3}\ln 3$