Calculate the following integrals using partial fractions: 3.1 integral x³ + (5-5x)/(6x²+x-1) dx, 3.2 integral (2x³+6x²-12)/(x(x+3)(x²+3x+4)) dx

The denominator is $6x^2+x-1$. We look for two numbers that multiply to $6 \times (-1) = -6$ and add to 1. These numbers are 3 and $-2$.

$$6x^2+x-1 = 6x^2+3x-2x-1 = 3x(2x+1) - 1(2x+1) = (3x-1)(2x+1)$$

$$\frac{5-5x}{(3x-1)(2x+1)} = \frac{A}{3x-1} + \frac{B}{2x+1}$$

Multiply both sides by $(3x-1)(2x+1)$:

$$5-5x = A(2x+1) + B(3x-1)$$

To find A, set $3x-1=0 \implies x=\frac{1}{3}$:

$$5-5\left(\frac{1}{3}\right) = A\left(2\left(\frac{1}{3}\right)+1\right) + B(0)$$ $$5-\frac{5}{3} = A\left(\frac{2}{3}+1\right)$$ $$\frac{15-5}{3} = A\left(\frac{5}{3}\right)$$ $$\frac{10}{3} = \frac{5}{3}A \implies A = 2$$

To find B, set $2x+1=0 \implies x=-\frac{1}{2}$:

$$5-5\left(-\frac{1}{2}\right) = A(0) + B\left(3\left(-\frac{1}{2}\right)-1\right)$$ $$5+\frac{5}{2} = B\left(-\frac{3}{2}-1\right)$$ $$\frac{10+5}{2} = B\left(-\frac{5}{2}\right)$$ $$\frac{15}{2} = -\frac{5}{2}B \implies B = -3$$

So, the partial fraction decomposition is:

$$\frac{5-5x}{6x^2+x-1} = \frac{2}{3x-1} - \frac{3}{2x+1}$$

$$\int \left(\frac{2}{3x-1} - \frac{3}{2x+1}\right) dx = \int \frac{2}{3x-1} dx - \int \frac{3}{2x+1} dx$$

For $\int \frac{2}{3x-1} dx$: Let $u=3x-1$, $du=3dx$.

$$\int \frac{2}{u} \frac{1}{3} du = \frac{2}{3} \int \frac{1}{u} du = \frac{2}{3} \ln|u| = \frac{2}{3} \ln|3x-1|$$

For $\int \frac{3}{2x+1} dx$: Let $v=2x+1$, $dv=2dx$.

$$\int \frac{3}{v} \frac{1}{2} dv = \frac{3}{2} \int \frac{1}{v} dv = \frac{3}{2} \ln|v| = \frac{3}{2} \ln|2x+1|$$

Combining these, we get:

$$\int \frac{5-5x}{6x^2+x-1} dx = \frac{2}{3} \ln|3x-1| - \frac{3}{2} \ln|2x+1| + C_2$$

$$\int x^3 + \frac{5-5x}{6x^2+x-1} dx = \frac{x^4}{4} + \frac{2}{3} \ln|3x-1| - \frac{3}{2} \ln|2x+1| + C$$

The final answer is $\boxed{\frac{x^4}{4} + \frac{2}{3} \ln|3x-1| - \frac{3}{2} \ln|2x+1| + C}$.