Determine the following integrals by means of partial fractions: (x3+5x2-3x-1)/((3x+2)2(x2-1)) dx and (8x3-x2+4x-3)/((3-2x)3) dx

The denominator is $(3x+2)^2(x^2-1) = (3x+2)^2(x-1)(x+1)$.
The partial fraction decomposition is:

Multiply both sides by the common denominator $(3x+2)^2(x-1)(x+1)$:

• Set $x = 1$: $1^3 + 5(1)^2 - 3(1) - 1 = C(3(1)+2)^2(1+1)$ $2 = C(5)^2(2) \implies 2 = 50C \implies C = \frac{2}{50} = \frac{1}{25}$

• Set $x = -1$: $(-1)^3 + 5(-1)^2 - 3(-1) - 1 = D(3(-1)+2)^2(-1-1)$ $-1 + 5 + 3 - 1 = D(-1)^2(-2) \implies 6 = -2D \implies D = -3$

• Set $x = -\frac{2}{3}$: $(-\frac{2}{3})^3 + 5(-\frac{2}{3})^2 - 3(-\frac{2}{3}) - 1 = B(-\frac{2}{3}-1)(-\frac{2}{3}+1)$ $-\frac{8}{27} + \frac{20}{9} + 2 - 1 = B(-\frac{5}{3})(\frac{1}{3})$ $\frac{-8 + 60 + 27}{27} = -\frac{5}{9}B \implies \frac{79}{27} = -\frac{5}{9}B \implies B = \frac{79}{27} \cdot (-\frac{9}{5}) = -\frac{79}{15}$

• Compare coefficients of $x^3$: The expanded equation is $x^3 + 5x^2 - 3x - 1 = A(3x^3 + 2x^2 - 3x - 2) + B(x^2-1) + C(9x^3+21x^2+16x+4) + D(9x^3+3x^2-8x-4)$. Comparing coefficients of $x^3$: $1 = 3A + 9C + 9D$ $1 = 3A + 9(\frac{1}{25}) + 9(-3)$ $1 = 3A + \frac{9}{25} - 27 \implies 1 = 3A + \frac{9 - 675}{25} \implies 1 = 3A - \frac{666}{25}$ $3A = 1 + \frac{666}{25} = \frac{25 + 666}{25} = \frac{691}{25} \implies A = \frac{691}{75}$

Let $u = 3-2x$. Then $2x = 3-u$, so $x = \frac{3-u}{2}$.
Also, $du = -2 dx$, which means $dx = -\frac{1}{2} du$.

Numerator: $8x^3 - x^2 + 4x - 3$

The integral becomes:

The constant term $\frac{3}{2}$ from $\frac{1}{2}(3-2x)$ can be absorbed into the integration constant $C$.
So, an equivalent form is: