Неодређени интеграли 4

Текст задатака објашњених у видео лекцији:

Пр.19)   Решити $\int {\frac{{dx}}{{\sqrt {3{x^2} - 3} }}} $

Пр.20)   $\int {\frac{{{x^2}dx}}{{1 + {x^2}}}} $

Пр.21)   $\int {\frac{{{x^3} - {x^2} + x}}{{{x^2} + 1}}} dx$

Пр.22)   $\int {\frac{{\left( {1 + 2{x^2}} \right)}}{{{x^2}\left( {1 + {x^2}} \right)}}} dx$

Пр.23)   $\int {\frac{{{x^4}dx}}{{{x^2} + 1}}} $

Пр.19)   $\int {\frac{{dx}}{{\sqrt {3{x^2} - 3} }}}  = \int {\frac{{dx}}{{\sqrt {3\left( {{x^2} - 1} \right)} }}}  = \int {\frac{{dx}}{{\sqrt 3  \cdot \sqrt {{x^2} - 1} }}}  = \frac{1}{{\sqrt 3 }}\int {\frac{{dx}}{{\sqrt {{x^2} - 1} }}}  =  $

$ = \frac{1}{{\sqrt 3 }}\ln \left| {x + \sqrt {{x^2} - 1} } \right| + C$

Пр.20)   $\int {\frac{{{x^2}dx}}{{1 + {x^2}}}}  = \int {\frac{{{x^2} + 1 - 1}}{{1 + {x^2}}}} dx = \int {\frac{{{x^2} + 1}}{{1 + {x^2}}}} dx - \int {\frac{1}{{1 + {x^2}}}} dx =$

$= \int {dx}  - \int {\frac{1}{{1 + {x^2}}}} dx = x - arctgx + C $

Пр.21)   $\int {\frac{{{x^3} - {x^2} + x}}{{{x^2} + 1}}} dx = \int {\frac{{{x^3} + x}}{{{x^2} + 1}}} dx - \int {\frac{{{x^2}}}{{{x^2} + 1}}} dx = \int {\frac{{x\left( {{x^2} + 1} \right)}}{{{x^2} + 1}}} dx - \int {\frac{{{x^2} + 1 - 1}}{{{x^2} + 1}}} dx = $

$ = \int x dx - \int {\frac{{{x^2} + 1}}{{{x^2} + 1}}} dx + \int {\frac{1}{{{x^2} + 1}}} dx = \frac{{{x^2}}}{2} - x + arctgx + C$

Пр.22)   $\int {\frac{{\left( {1 + 2{x^2}} \right)}}{{{x^2}\left( {1 + {x^2}} \right)}}} dx = \int {\frac{{1 + {x^2} + {x^2}}}{{{x^2}\left( {1 + {x^2}} \right)}}} dx = \int {\left( {\frac{{1 + {x^2}}}{{{x^2}\left( {1 + {x^2}} \right)}} + \frac{{{x^2}}}{{{x^2}\left( {1 + {x^2}} \right)}}} \right)} dx = $

$ = \int {\frac{{1 + {x^2}}}{{{x^2}\left( {1 + {x^2}} \right)}}} dx + \int {\frac{{{x^2}}}{{{x^2}\left( {1 + {x^2}} \right)}}} dx = \int {\frac{1}{{{x^2}}}} dx + \int {\frac{1}{{1 + {x^2}}}} dx = $

$ = \int {{x^{ - 2}}} dx + arctgx + C = \frac{{{x^{ - 1}}}}{{ - 1}} + arctgx + C =  - \frac{1}{x} + arctgx + C$

Пр.23)   $\int {\frac{{{x^4}dx}}{{{x^2} + 1}}}  = \int {\frac{{{x^4} - 1 + 1}}{{{x^2} + 1}}} dx = \int {\left( {\frac{{{x^4} - 1}}{{{x^2} + 1}} + \frac{1}{{{x^2} + 1}}} \right)} dx = \int {\frac{{{x^4} - 1}}{{{x^2} + 1}}} dx + \int {\frac{1}{{{x^2} + 1}}} dx = $

$ = \int {\frac{{\left( {{x^2} - 1} \right)\left( {{x^2} + 1} \right)}}{{{x^2} + 1}}} dx + \int {\frac{1}{{{x^2} + 1}}} dx = $

$ = \int {{x^2}} dx - \int {dx + \int {\frac{1}{{{x^2} + 1}}} dx = \frac{{{x^3}}}{3}}  - x + arctgx + C$