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Пр.1)   ${\int {\left( {5 - 2x} \right)} ^9}dx$

Пр.2)   $\int {\sqrt[3]{{4x + 3}}} dx$

Пр.3)   $\int {\frac{{xdx}}{{\sqrt {1 - {x^2}} }}} $

Пр.1)   ${\int {\left( {5 - 2x} \right)} ^9}dx=$

Смена: $5 - 2x = t/'$

${\left( {5 - 2x} \right)^\prime }dx = t'dt$

$ - 2dx = dt$

$dx = \frac{{dt}}{{ - 2}}$

$ = \int {{t^9}\frac{{dt}}{{ - 2}} =  - \frac{1}{2}} \int {{t^9}dt =  - \frac{1}{2}}  \cdot \frac{{{t^{10}}}}{{10}} + C =  - \frac{{{t^{10}}}}{{20}} + C =  - \frac{{{{\left( {5 - 2x} \right)}^{10}}}}{{20}} + C$

Пр.2)   $\int {\sqrt[3]{{4x + 3}}} dx=$

Смена: $4x + 3 = t/'$

${\left( {4x + 3} \right)^\prime }dx = t'dt$

$4dx = dt$

$dx = \frac{1}{4}dt$

$ = \int {\sqrt[3]{t}} \frac{1}{4}dt = \frac{1}{4}\int {\sqrt[3]{t}} dt = \frac{1}{4}\int {{t^{\frac{1}{3}}}} dt = \frac{1}{4} \cdot \frac{{{t^{\frac{4}{3}}}}}{{\frac{4}{3}}} + C = \frac{1}{4} \cdot \frac{3}{4}{t^{\frac{4}{3}}} + C = $

$= \frac{3}{{16}}\sqrt[3]{{{t^4}}} + C = \frac{3}{{16}}\sqrt[3]{{{{\left( {4x + 3} \right)}^4}}} + C$

Пр.3)   $\int {\frac{{xdx}}{{\sqrt {1 - {x^2}} }}} $

Смена: $1 - {x^2} = t/'$

${\left( {1 - {x^2}} \right)^\prime } = t'dt$

$ - 2xdx = dt$

$xdx = \frac{{dt}}{{ - 2}}$

$ = \int {\frac{{\frac{{dt}}{{ - 2}}}}{{\sqrt t }}}  =  - \frac{1}{2}\int {\frac{{dt}}{{\sqrt t }}}  =  - \frac{1}{2}\int {\frac{{dt}}{{{t^{\frac{1}{2}}}}}}  =  - \frac{1}{2}\int {{t^{ - \frac{1}{2}}}} dt = $

$ =  - \frac{1}{2}\frac{{{t^{ - \frac{1}{2} + 1}}}}{{ - \frac{1}{2} + 1}} + C =  - \frac{1}{2}\frac{{{t^{\frac{1}{2}}}}}{{\frac{1}{2}}} + C = \sqrt t  + C = \sqrt {1 - {x^2}}  + C$