Integral of 3x^2-(3/x^4)+3cbrt(x) dx

1. Integrate term-by-term:

   1. The integral of a constant times a function is the constant times the integral of the function:

      $$\int 3 \sqrt[3]{x}\, dx = 3 \int \sqrt[3]{x}\, dx$$

      1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$:

         $$\int \sqrt[3]{x}\, dx = \frac{3 x^{\frac{4}{3}}}{4}$$

      So, the result is: $\frac{9 x^{\frac{4}{3}}}{4}$

   1. Integrate term-by-term:

      1. The integral of a constant times a function is the constant times the integral of the function:

         $$\int 3 x^{2}\, dx = 3 \int x^{2}\, dx$$

         1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$:

            $$\int x^{2}\, dx = \frac{x^{3}}{3}$$

         So, the result is: $x^{3}$

      1. The integral of a constant times a function is the constant times the integral of the function:

         $$\int \left(- \frac{3}{x^{4}}\right)\, dx = - 3 \int \frac{1}{x^{4}}\, dx$$

         1. Don't know the steps in finding this integral.

            But the integral is

            $$- \frac{1}{3 x^{3}}$$

         So, the result is: $\frac{1}{x^{3}}$

      The result is: $x^{3} + \frac{1}{x^{3}}$

   The result is: $\frac{9 x^{\frac{4}{3}}}{4} + x^{3} + \frac{1}{x^{3}}$

2. Now simplify:

   $$\frac{\frac{9 x^{\frac{13}{3}}}{4} + x^{6} + 1}{x^{3}}$$

3. Add the constant of integration:

   $$\frac{\frac{9 x^{\frac{13}{3}}}{4} + x^{6} + 1}{x^{3}}+ \mathrm{constant}$$

The answer is:

$$\frac{\frac{9 x^{\frac{13}{3}}}{4} + x^{6} + 1}{x^{3}}+ \mathrm{constant}$$