Неодређени интеграли – метода смене 2

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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$