Интеграли. Особине интеграла. Метода смене. Једноставни примери.
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Пр.4) $\int {\frac{{{x^2}}}{{3{x^2} - 1}}} dx$
Пр.5) $\int {{x^3}} {\left( {2{x^4} - 3} \right)^7}dx$
Пр.6) $\int {x\sin \left( {2{x^2} - 1} \right)} dx$
Пр.7) $\int {x \cdot {e^{4{x^2} - 2}}} dx$
Пр.8) $\int {\frac{{\sqrt {1 - \ln x} }}{x}} dx$
Пр.9) $\int {\frac{1}{{x\sqrt[3]{{{{\ln }^2}x}}}}} dx$
Пр.4) $\int {\frac{{{x^2}}}{{3{x^2} - 1}}} dx=$
Смена: $3{x^3} - 1 = t/'$
${\left( {3{x^3} - 1} \right)^\prime }dx = t'dt$
$9{x^2}dx = dt$
${x^2}dx = \frac{{dt}}{9}$
$ = \int {\frac{{\frac{{dt}}{9}}}{t}} = \frac{1}{9}\int {\frac{{dt}}{t}} = \frac{1}{9}\ln t + C = \frac{1}{9}\ln \left| {3{x^3} - 1} \right| + C$
Пр.5) $\int {{x^3}} {\left( {2{x^4} - 3} \right)^7}dx=$
Смена: $2{x^4} - 3 = t/'$
$8{x^3}dx = dt$
${x^3}dx = \frac{{dt}}{8}$
$ = \int {{t^7}} \frac{{dt}}{8} = \frac{1}{8}\int {{t^7}dt = \frac{1}{8}} \frac{{{t^8}}}{8} + C = \frac{1}{{64}}{\left( {2{x^4} - 3} \right)^8} + C$
Пр.6) $\int {x\sin \left( {2{x^2} - 1} \right)} dx$
Смена: $2{x^2} - 1 = t/'$
$4xdx = dt$
$xdx = \frac{{dt}}{4}$
$ = \int {\sin t} \frac{{dt}}{4} = \frac{1}{4}\int {\sin tdt = - \frac{1}{4}} \cos t + C = - \frac{1}{4}\cos \left( {2{x^2} - 1} \right) + C$
Пр.7) $\int {x \cdot {e^{4{x^2} - 2}}} dx$
Смена: $4{x^2} - 2 = t/'$
$8xdx = dt$
$xdx = \frac{{dt}}{8}$
$= \int {{e^t}} \frac{{dt}}{8} = \frac{1}{8}\int {{e^t}dt = \frac{1}{8}} {e^t} + C = \frac{1}{8}{e^{4{x^2} - 2}} + C$
Пр.8) $\int {\frac{{\sqrt {1 - \ln x} }}{x}} dx$
Смена: $1 - \ln x = t/'$
$ - \frac{1}{x}dx = dt$
$\frac{{dx}}{x} = - dt$
$ = \int {\sqrt t } \left( { - dt} \right) = - \int {{t^{\frac{1}{2}}}dt = - \frac{{{t^{\frac{3}{2}}}}}{{\frac{3}{2}}}} + C = - \frac{2}{3}{\sqrt t ^3} + C = - \frac{2}{3}{\sqrt {1 - \ln x} ^3} + C$
Пр.9) $\int {\frac{1}{{x\sqrt[3]{{{{\ln }^2}x}}}}} dx$
Смена: $\ln x = t/'$
$\frac{1}{x}dx = dt$
$ = \int {\frac{1}{{\sqrt[3]{{{t^2}}}}}} dt = - \int {\frac{1}{{{t^{\frac{2}{3}}}}}dt = \int {{t^{ - \frac{2}{3}}}dt} } = - \frac{2}{3}{\sqrt t ^3} + C = $
$ = \frac{{{t^{ - \frac{2}{3} + 1}}}}{{ - \frac{2}{3} + 1}} + C = \frac{{{t^{\frac{1}{3}}}}}{{\frac{1}{3}}} + C = 3\sqrt[3]{t} + C = 3\sqrt[3]{{\ln x}} + C$