Question

Интеграл4(x^4-1)^2 x^3 dxспасибо заранее! :)

Answer 1

4x^3/sqrt(x^4+1)dx 
знак интеграла числмтель четыре умножить на икс в кубе, и всё это поделить на корень открывается икс в четвёртой степени плюс еденица ldx 
dx в знаменителе но под знак корня не попадает

Answer 2

Calculation of the Integral

To calculate the integral of the function 4(x^4-1)^2+x^3+dx, we can break it down into smaller parts and apply the power rule and the constant rule of integration.

The integral of 4(x^4-1)^2 can be calculated by expanding the expression and integrating each term separately. The integral of x^3 can be calculated using the power rule of integration.

Let's calculate the integral step by step:

1. Expand the expression (x^4-1)^2: - (x^4-1)^2 = (x^4)^2 - 2(x^4)(1) + (1)^2 = x^8 - 2x^4 + 1

2. Calculate the integral of 4(x^4-1)^2: - ∫4(x^4-1)^2 dx = 4∫(x^8 - 2x^4 + 1) dx - Apply the power rule of integration: - ∫x^n dx = (x^(n+1))/(n+1) + C - Applying the power rule to each term: - 4∫(x^8 - 2x^4 + 1) dx = 4((x^9)/9 - (2x^5)/5 + x) + C - Simplifying the expression: - = (4/9)x^9 - (8/5)x^5 + 4x + C

3. Calculate the integral of x^3: - ∫x^3 dx = (x^4)/4 + C

4. Combine the results: - ∫4(x^4-1)^2+x^3+dx = (4/9)x^9 - (8/5)x^5 + 4x + (x^4)/4 + C - Simplifying the expression: - = (4/9)x^9 - (8/5)x^5 + (1/4)x^4 + 4x + C

Therefore, the integral of 4(x^4-1)^2+x^3+dx is (4/9)x^9 - (8/5)x^5 + (1/4)x^4 + 4x + C.

Please note that the constant of integration C represents an arbitrary constant and can be added to the result.