Set up the partial fraction decomposition.

The denominator has a linear factor $(x-3)$ and an irreducible quadratic factor $(x^2+9)$.

$$\frac{x^2}{(x^2+9)(x-3)} = \frac{A}{x-3} + \frac{Bx+C}{x^2+9}$$

Substitute $x=3$:

$$3^2 = A(3^2+9) + (B(3)+C)(3-3)$$ $$9 = A(9+9) + 0$$ $$9 = 18A$$ $$A = \frac{9}{18} = \frac{1}{2}$$

Expand the equation:

$$x^2 = Ax^2 + 9A + Bx^2 - 3Bx + Cx - 3C$$ $$x^2 = (A+B)x^2 + (-3B+C)x + (9A-3C)$$

Equate coefficients:
Coefficient of $x^2$:

$$1 = A+B$$

Substitute $A=\frac{1}{2}$:

$$1 = \frac{1}{2} + B$$ $$B = 1 - \frac{1}{2} = \frac{1}{2}$$

Coefficient of $x$:

$$0 = -3B+C$$

Substitute $B=\frac{1}{2}$:

$$0 = -3\left(\frac{1}{2}\right) + C$$ $$C = \frac{3}{2}$$

Constant term (for verification):

$$0 = 9A-3C$$

Substitute $A=\frac{1}{2}$ and $C=\frac{3}{2}$:

$$0 = 9\left(\frac{1}{2}\right) - 3\left(\frac{3}{2}\right)$$ $$0 = \frac{9}{2} - \frac{9}{2}$$ $$0 = 0$$

The values are consistent.

$$\frac{x^2}{(x^2+9)(x-3)} = \frac{\frac{1}{2}}{x-3} + \frac{\frac{1}{2}x+\frac{3}{2}}{x^2+9}$$ $$\frac{x^2}{(x^2+9)(x-3)} = \frac{1}{2(x-3)} + \frac{x+3}{2(x^2+9)}$$

The final answer is $\boxed{\frac{1}{2(x-3)} + \frac{x+3}{2(x^2+9)}}$.

The denominator has a linear factor $(x-1)$ and an irreducible quadratic factor $(x^2+4)$.

$$\frac{1}{(x-1)(x^2+4)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+4}$$

Substitute $x=1$:

$$1 = A(1^2+4) + (B(1)+C)(1-1)$$ $$1 = A(1+4) + 0$$ $$1 = 5A$$ $$A = \frac{1}{5}$$

Expand the equation:

$$1 = Ax^2 + 4A + Bx^2 - Bx + Cx - C$$ $$1 = (A+B)x^2 + (-B+C)x + (4A-C)$$

Equate coefficients:
Coefficient of $x^2$:

$$0 = A+B$$

Substitute $A=\frac{1}{5}$:

$$0 = \frac{1}{5} + B$$ $$B = -\frac{1}{5}$$

Coefficient of $x$:

$$0 = -B+C$$

Substitute $B=-\frac{1}{5}$:

$$0 = -\left(-\frac{1}{5}\right) + C$$ $$0 = \frac{1}{5} + C$$ $$C = -\frac{1}{5}$$

Constant term (for verification):

$$1 = 4A-C$$

Substitute $A=\frac{1}{5}$ and $C=-\frac{1}{5}$:

$$1 = 4\left(\frac{1}{5}\right) - \left(-\frac{1}{5}\right)$$ $$1 = \frac{4}{5} + \frac{1}{5}$$ $$1 = \frac{5}{5}$$ $$1 = 1$$

The values are consistent.

$$\frac{1}{(x-1)(x^2+4)} = \frac{\frac{1}{5}}{x-1} + \frac{-\frac{1}{5}x-\frac{1}{5}}{x^2+4}$$ $$\frac{1}{(x-1)(x^2+4)} = \frac{1}{5(x-1)} - \frac{x+1}{5(x^2+4)}$$

The final answer is $\boxed{\frac{1}{5(x-1)} - \frac{x+1}{5(x^2+4)}}$.

To find $B$, substitute $x=-1$:

$$6(-1)+7 = A(-1+1) + B(-1+3)$$ $$-6+7 = A(0) + B(2)$$ $$1 = 2B$$ $$B = \frac{1}{2}$$

To find $A$, substitute $x=-3$:

$$6(-3)+7 = A(-3+1) + B(-3+3)$$ $$-18+7 = A(-2) + B(0)$$ $$-11 = -2A$$ $$A = \frac{-11}{-2} = \frac{11}{2}$$

$$B-A = \frac{1}{2} - \frac{11}{2}$$ $$B-A = \frac{1-11}{2}$$ $$B-A = \frac{-10}{2}$$ $$B-A = -5$$

The final answer is $\boxed{-5}$.