Determine the following integrals by means of partial fractions: 3.1 ∫(5x³ - x² + 8x - 4) / ((1-x)² (2x + 1)²) dx 3.2 ∫(4x³ - 6x² + 5x - 4) / (2x² + x - 6) dx

Step 1: Perform polynomial long division for integral 3.2.
For integral 3.2, the degree of the numerator ($4x^3 - 6x^2 + 5x - 4$) is 3, and the degree of the denominator ($2x^2 + x - 6$) is 2. Since the numerator's degree is greater than or equal to the denominator's degree, we must perform polynomial long division first.

$$\begin{array}{c|cc cc} \multicolumn{2}{r}{2x} & -4 \\ \cline{2-5} 2x^2+x-6 & 4x^3 & -6x^2 & +5x & -4 \\ \multicolumn{2}{r}{-(4x^3} & +2x^2 & -12x) \\ \cline{2-4} \multicolumn{2}{r}{0} & -8x^2 & +17x & -4 \\ \multicolumn{2}{r}{} & -(-8x^2 & -4x & +24) \\ \cline{3-5} \multicolumn{2}{r}{} & 0 & 21x & -28 \\ \end{array}$$

So, $\frac{4x^3 - 6x^2 + 5x - 4}{2x^2 + x - 6} = 2x - 4 + \frac{21x - 28}{2x^2 + x - 6}$.

Step 2: Factor the denominator for integral 3.2.
The denominator is $2x^2 + x - 6$. We find the roots using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

$$x = \frac{-1 \pm \sqrt{1^2 - 4(2)(-6)}}{2(2)} = \frac{-1 \pm \sqrt{1 + 48}}{4} = \frac{-1 \pm \sqrt{49}}{4} = \frac{-1 \pm 7}{4}$$

The roots are $x_1 = \frac{-1+7}{4} = \frac{6}{4} = \frac{3}{2}$ and $x_2 = \frac{-1-7}{4} = \frac{-8}{4} = -2$.
So, $2x^2 + x - 6 = 2(x - \frac{3}{2})(x + 2) = (2x - 3)(x + 2)$.

Step 3: Decompose the remainder term into partial fractions for integral 3.2.
We set up the partial fraction decomposition for $\frac{21x - 28}{(2x - 3)(x + 2)}$:

$$\frac{21x - 28}{(2x - 3)(x + 2)} = \frac{A}{2x - 3} + \frac{B}{x + 2}$$

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

$$21x - 28 = A(x + 2) + B(2x - 3)$$

To find $A$, set $x = \frac{3}{2}$:

$$21\left(\frac{3}{2}\right) - 28 = A\left(\frac{3}{2} + 2\right) + B(0)$$ $$\frac{63}{2} - \frac{56}{2} = A\left(\frac{7}{2}\right)$$ $$$$