To evaluate the integral

$\int_{3}^{-4} \frac{3}{x^4} \, dx,$

we first recognize that the integral has limits of integration from $3$ to $-4$. However, typically, the limits of integration should go from a lower value to a higher value. So we will first rewrite the integral by swapping the limits and changing the sign:

$\int_{3}^{-4} \frac{3}{x^4} \, dx = -\int_{-4}^{3} \frac{3}{x^4} \, dx.$

Now, let's evaluate the integral:

$\int \frac{3}{x^4} \, dx.$

We can rewrite $\frac{3}{x^4}$ as $3x^{-4}$. The antiderivative of $x^{-4}$ is given by:

$\int x^{-4} \, dx = \frac{x^{-3}}{-3} = -\frac{1}{3} x^{-3}.$

Therefore, the antiderivative of $3x^{-4}$ is:

$3 \left( -\frac{1}{3} x^{-3} \right) = -x^{-3} = -\frac{1}{x^3}.$

So the integral becomes:

[ -\int_{-4}^{3} \frac{3}{x^4} , dx = -\left[ -\frac{1}{x^3} \right]{-4}^{3} = \left[ \frac{1}{x^3} \right]{-4}^{3}. ]

Now we evaluate this from $-4$ to $3$:

$\left[ \frac{1}{x^3} \right]_{-4}^{3} = \frac{1}{3^3} - \frac{1}{(-4)^3} = \frac{1}{27} - \frac{1}{-64}.$

Simplifying this, we get:

$\frac{1}{27} + \frac{1}{64}.$

To combine these fractions, we find a common denominator, which is $27 \times 64 = 1728$:

$\frac{1}{27} = \frac{64}{1728}, \quad \frac{1}{64} = \frac{27}{1728}.$

Adding these together:

$\frac{64}{1728} + \frac{27}{1728} = \frac{91}{1728}.$

Thus, the value of the integral is:

$\boxed{\frac{91}{1728}}.$

Do you want more details or have any questions?

Here are some more questions you might want to explore:

1. Evaluate $\int_1^2 \frac{1}{x} \, dx$.
2. Find the derivative of $f(x) = x^2 \ln x$.
3. Determine the convergence or divergence of $\sum_{n=1}^\infty \frac{1}{n^3}$.
4. Solve the differential equation $\frac{dy}{dx} = e^x \cos y$.
5. Compute the integral $\int_0^{\pi/2} \sin x \, dx$.

Tip: Always ensure the limits of integration are in the correct order; swap them and change the sign if needed to avoid confusion.