Question

Miroslava227 , помогите пожалуйста с математикой ,интегралы

Answer 1

Ответ:

1

$\int\limits(6x - \frac{2 {x}^{3} \sqrt{x} }{ \sqrt[3]{x} } )dx = \frac{6 {x}^{2} }{2} - \int\limits2 {x}^{3 + \frac{1}{2} - \frac{1}{3} } dx = \\ = 3 {x}^{2} - 2\int\limits {x}^{ \frac{19}{6} } dx = 3 {x}^{2} - 2 \times \frac{ {x}^{ \frac{25}{6} } }{ \frac{25}{6} } + C = \\ = 3 {x}^{2} - \frac{12}{25} {x}^{4} \sqrt[6]{x} + C$

2

$\int\limits \frac{ \sqrt{ {x}^{2} + 5} }{6} xdx \\ \\ {x}^{2} + 5 = t \\ 2xdx = dt \\ xdx = \frac{dt}{2} \\ \\ \frac{1}{2} \times \frac{1}{6} \int\limits \sqrt{t} dt = \frac{1}{12} \times \frac{ {t}^{ \frac{3}{2} } }{ \frac{3}{2} } + C= \\ = \frac{1}{18} t\sqrt{t} + C = \frac{1}{18} \sqrt{ {( {x}^{2} + 5)}^{3} } + C$

3

$\int\limits {32}^{x} \times xdx \\ \\ u = x \: \: \: du = dx \\ dv = {32}^{x} dx \: \: \: \: v = \frac{ {32}^{x} }{ ln(32) } \\ \\ \frac{ {32}^{x}x }{ ln(32) } - \frac{1}{ ln(32) } \int\limits {32}^{x} dx = \\ = \frac{ {32}^{x}x }{ ln(32) } - \frac{1}{ ln(32) } \times \frac{ {32}^{x} }{ ln(32) } + C= \\ = \frac{ {32}^{x} }{ ln(32) } (x - \frac{1}{ ln(32) } )+ C$

4

$\int\limits \sin {}^{3} (x) \cos(x) dx = \int\limits \sin {}^{3} (x) d( \sin(x)) = \\ = \frac{ \sin {}^{4} (x) }{4} + C$

5

$\int\limits \frac{dx}{6x + 5} = \frac{1}{6} \int\limits \frac{d(6x + 5)}{6x + 5} = \frac{1}{6} ln( |6x + 5| ) + C$