Question

Найдите первообразные следующих функций:

№1

1) f(x) = 3x; 2) f(x) = 4x2 + x – 2; 3) f(x) = + 1; 4) f(x) =

№2.

1. f(x) = 2sinx; 2) f(x) = 5cosx; 3) f(x) = 3cosx - 4sinx;

4)f(x) = 5sinx + 2 cosx; 5) f(x) = x2 + ; 6) f(x) = x3 - ; 7) f(x) = sin(3x + ); 8) f(x) = cos(2x + ).

Answer 1

Ответ:

1.

$1)∫3xdx = \frac{3 {x}^{2} }{2} + c$

$2)∫(4 {x}^{2} + x - 2)dx = \frac{4 {x}^{3} }{3} + \frac{ {x}^{2} }{2} - 2x + c$

$3)∫( \frac{ {x}^{3} }{3} + 1)dx = \frac{ {x}^{4} }{12} + x + c$

$4)∫ {x}^{ - 2} dx = \frac{ {x}^{ - 1} }{ - 1} + c = - \frac{1}{x} + c$

2.

$1)∫2 \sin(x) dx = - 2\cos(x) + c$

$2)∫5 \cos(x) dx = 5 \sin(x) + c$

$3)∫(3 \cos(x) - 4 \sin(x) )dx = 3 \sin(x) + 4 \cos(x) + c$

$4)∫(5 \sin(x) + 2 \cos(x) )dx = - 5 \cos(x) + 2 \sin(x) + c$

$5)∫( {x}^{2} + 3 {x}^{ - \frac{1}{2} } )dx = \frac{ {x}^{3} }{3} + 3 \frac{ {x}^{ \frac{1}{2} } }{ \frac{1}{2} } + c = \frac{ {x}^{3} }{3} + 6 \sqrt{x} + c$

$6)∫( {x}^{3} - 4 {x}^{ - \frac{1}{2} } )dx = \frac{ {x}^{4} }{4} - 8 \sqrt{x} + c$

$7)∫ \sin( x+ \frac{\pi}{3} ) dx = ∫ \sin(x + \frac{\pi}{3} ) d(x + \frac{\pi}{3} = = - \cos(x + \frac{\pi}{3} ) + c$

$8)∫ \cos(2x + \frac{\pi}{6} ) dx = \frac{1}{2} ∫ \cos(2x + \frac{\pi}{6} ) d(2x + \frac{\pi}{6} ) = \frac{1}{2} \sin(2x + \frac{\pi}{6} ) + c$