Find \[ \int \frac{3 x^{4}-4}{2 x^{3}} \mathrm{~d} x \] writing your answer in simplest form.

To solve the integral \[ \int \frac{3 x^{4}-4}{2 x^{3}} \, \mathrm{d}x, \] we can first simplify the integrand. We can break it into two separate fractions: \[ \frac{3 x^{4}}{2 x^{3}} - \frac{4}{2 x^{3}} = \frac{3}{2} x - \frac{2}{x^{3}}. \] Now, we can write the integral as: \[ \int \left( \frac{3}{2} x - \frac{2}{x^{3}} \right) \, \mathrm{d}x. \] Next, we can integrate each term separately: 1. For \(\int \frac{3}{2} x \, \mathrm{d}x\), we use the power rule: \[ \int \frac{3}{2} x \, \mathrm{d}x = \frac{3}{2} \cdot \frac{x^2}{2} = \frac{3}{4} x^2. \] 2. For \(\int -\frac{2}{x^{3}} \, \mathrm{d}x\), we rewrite it as \(-2 \int x^{-3} \, \mathrm{d}x\): \[ -2 \int x^{-3} \, \mathrm{d}x = -2 \cdot \left( \frac{x^{-2}}{-2} \right) = x^{-2} = \frac{1}{x^{2}}. \] Combining these results, we get: \[ \int \frac{3 x^{4}-4}{2 x^{3}} \, \mathrm{d}x = \frac{3}{4} x^{2} + \frac{1}{x^{2}} + C, \] where \(C\) is the constant of integration. Thus, the final answer is: \[ \frac{3}{4} x^{2} + \frac{1}{x^{2}} + C. \]