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

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Calculate the integral $$ \int \frac{3x^4-4}{2x^3}dx $$.
Evaluate the integral by following steps:
- step0: Evaluate using formulas and rules:
  $$\int \frac{3x^{4}-4}{2x^{3}} dx$$
- step1: Use properties of integrals:
  $$\frac{1}{2}	imes \int \frac{3x^{4}-4}{x^{3}} dx$$
- step2: Rewrite the fraction:
  $$\frac{1}{2}	imes \int 3x-\frac{4}{x^{3}} dx$$
- step3: Use properties of integrals:
  $$\frac{1}{2}\left(\int 3x dx+\int -\frac{4}{x^{3}} dxight)$$
- step4: Calculate:
  $$\frac{1}{2}	imes \int 3x dx+\frac{1}{2}	imes \int -\frac{4}{x^{3}} dx$$
- step5: Evaluate the integral:
  $$\frac{3x^{2}}{4}+\frac{1}{2}	imes \int -\frac{4}{x^{3}} dx$$
- step6: Evaluate the integral:
  $$\frac{3x^{2}}{4}+\frac{1}{x^{2}}$$
- step7: Add the constant of integral C:
  $$\frac{3x^{2}}{4}+\frac{1}{x^{2}} + C, C \in \mathbb{R}$$
The integral of $$\frac{3x^{4}-4}{2x^{3}}$$ with respect to $$x$$ is $$\frac{3x^{2}}{4} + \frac{1}{x^{2}} + C$$, where $$C$$ is an arbitrary constant.
The integral simplifies to $$ \frac{3x^{2}}{4} + \frac{1}{x^{2}} + C $$.

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