1) ∫ (2x^2 - 1 / (x^2 + 5) dx = 2) ∫ (x-2) cos3x dx = 3) ∫ (4x^3 + x^2 + 2) / (x (x-1) (x-2)) 4) ∫ dx / (cosx (1+cosx)) = 5) ∫ sin3x * 4^cos3x+1 dx =

Question

Ярослав Попов

Математика

9 января, 02:00

1) ∫ (2x^2 - 1 / (x^2 + 5) dx = 2) ∫ (x-2) cos3x dx = 3) ∫ (4x^3 + x^2 + 2) / (x (x-1) (x-2)) 4) ∫ dx / (cosx (1+cosx)) = 5) ∫ sin3x * 4^cos3x+1 dx =

+1

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Платон Яковлев · Answer 1

1) f (x) ' = (x * (x - 4)) ' = (x) ' * (x - 4) + x * (x - 4) ' = (x) ' * (x - 4) + x * ((x) ' - (4) ') = 1 * (x - 4) + x * (1 - 0) = x - 4 + x = 2x - 4.

2) f (x) ' = (3 - 2 х) ' = (3) ' - (2 х) ' = 0 - 2 * 1 * x^{(1 - 1)} = - 2 * x⁰ = - 2 * 1 = - 2.

3) f (x) ' = ((-x^2 + 2x^2) ^3 + (x - 3) ^4) ' = ((-x^2 + 2x^2) ^3) ' + ((x - 3) ^4) ' = (-x^2 + 2x^2) ' * ((-x^2 + 2x^2) ^3) ' + (x - 3) ' * ((x - 3) ^4) ' = ((-x^2) ' + (2x^2) ') * ((-x^2 + 2x^2) ^3) ' + ((x) ' - (3) ') * ((x - 3) ^4) ' = (-2x + 4x) * (3 * (-x^2 + 2x^2) ^2) + (1 - 0) * (4 * (x - 3) ^3) = 2x * 3 * (x^2) ^2 + 1 * (4 * (x - 3) ^3) = 6x * x^4 + 4 (x - 3) ^3 = 6x^5 + 4 (x - 3) ^3.

4) f (x) ' = (x^2 * tg (x)) ' = (x^2) ' * tg (x) + x^2 * (tg (x)) ' = 2x * tg (x) + x^2 * (1 / (cos^2 (x))) = 2xtg (x) + x^2 / (cos^2 (x)).

5) f (x) ' = (sin^2 (x)) ' = (sin (x)) ' * (sin^2 (x)) ' = cos (x) * 2 * sin^ (2 - 1) (x) = cos (x) * 2 * sin^ (1) (x) = cos (x) * 2 * sin (x) = 2cos (x) sin (x).