To solve the integral $\int \frac{2x^4 - 3}{x^4} \, dx$:

Step 1: Simplify the expression

The integrand $\frac{2x^4 - 3}{x^4}$ can be split as: $\frac{2x^4 - 3}{x^4} = \frac{2x^4}{x^4} - \frac{3}{x^4} = 2 - \frac{3}{x^4}.$

Step 2: Rewrite the integral

The integral becomes: $\int \left( 2 - \frac{3}{x^4} \right) dx = \int 2 \, dx - \int \frac{3}{x^4} \, dx.$

Step 3: Solve each part

1. For $\int 2 \, dx$: $\int 2 \, dx = 2x.$

2. For $\int \frac{3}{x^4} \, dx$: Rewrite $\frac{3}{x^4}$ as $3x^{-4}$. Then: $\int 3x^{-4} \, dx = 3 \cdot \frac{x^{-4+1}}{-4+1} = 3 \cdot \frac{x^{-3}}{-3} = -\frac{3}{x^3}.$

Step 4: Combine results

Adding the results together: $\int \frac{2x^4 - 3}{x^4} \, dx = 2x - \frac{3}{x^3} + C,$ where $C$ is the constant of integration.

Final Answer:

The correct option is Ответ 2: $2x - \frac{3}{x^3} + C$.

Would you like further clarification or a step-by-step derivation of any part?

Related Questions:

1. What is the general approach for splitting fractions in integrals?
2. How do you integrate terms with negative exponents?
3. What is the importance of adding the constant of integration $C$?
4. Can this problem be solved using substitution or other methods?
5. How can the result be verified using differentiation?

Tip:

Always simplify the integrand before integrating—it makes the process more manageable!