Question

помогитее очень прошу​

Answer 1

Ответ:

13.

$\int\limits \sin {}^{2} ( \frac{x}{2} ) dx \\ \\ \sin {}^{2} ( \frac{x}{2} ) = \frac{1}{2}( 1 - \cos(x) ) \\ \\ \int\limits \frac{1}{2} (1 - \cos(x)) dx = \frac{1}{2} \int\limits(1- \cos(x)) dx = \\ = \frac{1}{2} (x - \sin(x)) + C$

14.

$\int\limits \frac{ \sqrt{1 + {x}^{2} } - \sqrt{1 - {x}^{2} } }{ \sqrt{1 - {x}^{4} } } dx = \int\limits( \frac{ \sqrt{1 + {x}^{2} } }{ \sqrt{1 - {x}^{4} } } - \frac{ \sqrt{1 - {x}^{2} } }{ \sqrt{1 - {x}^{4} } } )dx = \\ = \int\limits( \frac{ \sqrt{1 + {x}^{2} } }{ \sqrt{(1 + {x}^{2})(1 - {x}^{2}) } } - \frac{ \sqrt{1 - {x}^{2} } }{ \sqrt{(1 - {x}^{2} )(1 + {x}^{2}) } } )dx = \\ = \int\limits( \frac{1}{ \sqrt{1 - {x}^{2} } } - \frac{1}{ \sqrt{1 - {x}^{2} } } )dx = \\ = arcsin(x) - ln( |x + \sqrt{1 - {x}^{2} } | ) + C$

15.

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

16.

$\int\limits( \frac{1}{ {x}^{2} - 25} + \frac{1}{ \sqrt{ {x}^{2} + 5} } )dx = \\ = \int\limits \frac{dx}{ {x}^{2} - {5}^{2} } + \int\limits \frac{dx}{ \sqrt{ {x}^{2} + {( \sqrt{5} )}^{2} } } = \\ = \frac{1}{2 \times 5} ln( | \frac{x - 5}{x + 5} | ) + ln( |x + \sqrt{ {x}^{2} + 5 } | ) + C \\ = \frac{1}{10} ln( | \frac{x - 5}{x + 5} | ) + ln( |x + \sqrt{ {x}^{2} + 5 } | ) + C$

17.

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

18.

$\int\limits \frac{ {x}^{2} + 2 }{ {x}^{2} - 1} dx= \int\limits \frac{ {x}^{2} - 1 + 3 }{ {x}^{2} - 1} dx = \\ = \int\limits(1 + \frac{3}{ {x}^{2} - 1} )dx = x + \frac{3}{2} ln( | \frac{x - 1}{x + 1} | ) + C$