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Неодређени интеграли - метода смене 5

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$