Question

Помогите решить пожалуйста

Answer 1

Ответ:

1

$\int\limits \: d(tgx) = tgx + C$

2

$\int\limits {e}^{8x} dx = \frac{1}{8}\int\limits {e}^{8x} d(8x) = \frac{1}{8} {e}^{8x} + C$

3

$\int\limits {(7x + 3)}^{8} dx = \frac{1}{7} \int\limits {(7x + 3)}^{8} d(7x) = \\ = \frac{1}{7} \int\limits {(7x + 3)}^{8} d(7x + 3) = \\ = \frac{ {(7x + 3)}^{9} }{63} + C$

4

$\int\limits \frac{dx}{ {(x - 3)}^{2} + 16} = \int\limits \frac{d(x - 3)}{ {(x - 3)}^{2} + {4}^{2} } = \\ = \frac{1}{4} arctg( \frac{x - 3}{4} ) + C$

5

$\int\limits \cos(7x) dx= \frac{1}{7}\int\limits \cos(7x) d(7x) = \\ = \frac{1}{7} \sin(7x) + C$

6

$\int\limits \frac{dx}{ \sqrt{7x + 1} } = \frac{1}{7} \int\limits {(7x + 1)}^{ - \frac{1}{2} } d(7x + 1) = \\ = \frac{1}{7} \times \frac{ {(7x + 1)}^{ \frac{1}{2} } }{ \frac{1}{2} } + C = \frac{2}{7} \sqrt{7x + 1} + C$

7

$\int\limits {x}^{3} \sin( {x}^{4} ) dx = \frac{1}{4} \int\limits {x}^{3} \sin( {x}^{4} ) dx = \\ = \frac{1}{4} \int\limits\sin( {x}^{4} ) d( {x}^{4} ) = - \frac{1}{4} \cos( {x}^{4} ) + C$

8

$\int\limits\frac{d( {e}^{x}) }{ {( {e}^{x} )}^{2} - {2}^{2} } = \frac{1}{2 \times 2} ln( \frac{ {e}^{x} - 2}{ {e}^{x} + 2} ) + c = \\ = \frac{1}{4} ln( \frac{ {e}^{x} - 2 }{ {e}^{x} + 2} ) + C$

9

$\int\limits \frac{d( {x}^{3}) }{ \sqrt{ {3}^{2} - {( {x}^{3} )}^{2} } } = arcsin( \frac{ {x}^ {3} }{3} ) + C$

10

$\int\limits \frac{d( \cos(x)) }{ \cos(x) } = ln( \cos(x) ) + C$

11

$\int\limits {5}^{3x} dx = \frac{1}{3} \int\limits {5}^{3x} d(3x) = \frac{ {5}^{3x} }{ 3ln(5) } + C$

12

$\int\limits \frac{dx}{ { \cos }^{2}5x } = \frac{1}{5} \int\limits \frac{d(5x)}{ { \cos}^{2}(5x) } = \frac{1}{5} tg5x + C$