Интеграли. Особине интеграла. Метода смене. Сложенији примери.
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Пр.17) $\int {\frac{{dx}}{{1 + 9{x^2}}}} $
Пр.18) $\int {\frac{{dx}}{{\sqrt {4 - 5{x^2}} }}} $
Пр.19) $\int {\frac{{dx}}{{{{\left( {3x - 5} \right)}^2} - 4}}} $
Пр.20) $\int {\frac{{dx}}{{{x^2} - 4x - 4}}} $
Пр.21) $\int {\frac{{dx}}{{{x^2} + x + 1}}} $
Пр.17) $\int {\frac{{dx}}{{1 + 9{x^2}}}} = \int {\frac{{dx}}{{1 + {{\left( {3x} \right)}^2}}}} = $
Смена: $3x = t|'$
$3dx = dt$
$dx = \frac{{dt}}{3}$
$ = \int {\frac{{\frac{{dt}}{3}}}{{1 + {{\left( t \right)}^2}}}} = \frac{1}{3}\int {\frac{{dt}}{{1 + {t^2}}}} = \frac{1}{3}arctgt + C = \frac{1}{3}arctg3x + C$
Пр.18) $\int {\frac{{dx}}{{\sqrt {4 - 5{x^2}} }}} = \int {\frac{{dx}}{{\sqrt {4\left( {1 - \frac{5}{4}{x^2}} \right)} }}} = \int {\frac{{dx}}{{\sqrt 4 \sqrt {1 - {{\left( {\frac{{\sqrt 5 }}{2}x} \right)}^2}} }}} = $
Смена: $\frac{{\sqrt 5 }}{2}x = t$
$\frac{{\sqrt 5 }}{2}dx = dt$
$dx = \frac{2}{{\sqrt 5 }}dt$
$ = \frac{1}{2}\int {\frac{{\frac{2}{{\sqrt 5 }}dt}}{{\sqrt {1 - {t^2}} }}} = \frac{1}{2} \cdot \frac{2}{{\sqrt 5 }}\int {\frac{{dt}}{{\sqrt {1 - {t^2}} }}} = \frac{1}{{\sqrt 5 }}\arcsin t + C = \frac{{\sqrt 5 }}{5}\arcsin \frac{{\sqrt 5 }}{2}x + C$
Пр.19) $\int {\frac{{dx}}{{{{\left( {3x - 5} \right)}^2} - 4}}} = \int {\frac{{dx}}{{ - 4\left( {1 - \frac{{{{\left( {3x - 5} \right)}^2}}}{4}} \right)}}} = - \frac{1}{4}\int {\frac{{dx}}{{1 - {{\left( {\frac{{3x - 5}}{2}} \right)}^2}}}} = $
Смена: $\frac{{3x - 5}}{2} = t$
${\left( {\frac{{3x}}{2} - \frac{5}{2}} \right)^\prime }dx = t'dt$
$\frac{3}{2}dx = dt$
$dx = \frac{2}{3}dt$
$ = - \frac{1}{4}\int {\frac{{\frac{2}{3}dt}}{{1 - {t^2}}}} = - \frac{1}{4} \cdot \frac{2}{3}\int {\frac{{dt}}{{1 - {t^2}}}} = - \frac{1}{6} \cdot \frac{1}{2}\ln \left| {\frac{{1 + t}}{{1 - t}}} \right| + C = $
$ = - \frac{1}{{12}}\ln \left| {\frac{{1 + \frac{{3x - 5}}{2}}}{{1 - \frac{{3x - 5}}{2}}}} \right| + C = - \frac{1}{{12}}\ln \left| {\frac{{\frac{{2 + 3x - 5}}{2}}}{{\frac{{2 - 3x + 5}}{2}}}} \right| + C = $
$ = - \frac{1}{{12}}\ln \left| {\frac{{3x - 3}}{{7 - 3x}}} \right| + C$
Пр.20) $\int {\frac{{dx}}{{{x^2} - 4x - 4}}} $
Смена: $x - 2 = t$
$\frac{{\sqrt 5 }}{2}dx = dt$
$dx = dt$
$ = \int {\frac{{dt}}{{{t^2}}}} = \int {{t^{ - 2}}dt = \frac{{{t^{ - 3}}}}{{ - 3}}} + C = - \frac{1}{3} \cdot \frac{1}{{{t^3}}} + C = - \frac{1}{{3\left( {x - 2} \right)}} + C$
Пр.21) $\int {\frac{{dx}}{{{x^2} + x + 1}}} = \int {\frac{{dx}}{{{x^2} + x + \frac{1}{4} - \frac{1}{4} + 1}}} = \int {\frac{{dx}}{{{{\left( {x + \frac{1}{2}} \right)}^2} + \frac{3}{4}}}} = \int {\frac{{dx}}{{\frac{3}{4}\left( {1 + \frac{{{{\left( {x + \frac{1}{2}} \right)}^2}}}{{\frac{3}{4}}}} \right)}} = } $
$ = \frac{4}{3}\int {\frac{{dx}}{{1 + \frac{{4{{\left( {x + \frac{1}{2}} \right)}^2}}}{3}}} = } \frac{4}{3}\int {\frac{{dx}}{{1 + {{\left( {\frac{{2\left( {x + \frac{1}{2}} \right)}}{{\sqrt 3 }}} \right)}^2}}} = } $
Смена: $\frac{{2x + 1}}{{\sqrt 3 }} = t$
${\left( {\frac{{2x + 1}}{{\sqrt 3 }}} \right)^\prime }dx = t'dt$
$\frac{2}{{\sqrt 3 }}dx = dt$
$dx = \frac{{\sqrt 3 }}{2}dt$
$ = \frac{4}{3}\int {\frac{{\frac{{\sqrt 3 }}{2}dt}}{{1 + {t^2}}} = } \frac{4}{3} \cdot \frac{{\sqrt 3 }}{2}\int {\frac{{dt}}{{1 + {t^2}}} = } \frac{{2\sqrt 3 }}{3}arctgt + C = \frac{{2\sqrt 3 }}{3}arctg\frac{{2x + 1}}{{\sqrt 3 }} + C$