Интеграли. Особине интеграла. Метода смене. Сложенији примери.
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Пр.22) Решити $\int {\frac{{dx}}{{\sqrt {1 - {x^2}} {{\arcsin }^4}x}}} $
Пр.23) $\int {\frac{{x - arctgx}}{{1 + {x^2}}}} dx$
Пр.24) $\int {\frac{{\ln 3x}}{{x\ln 9x}}} dx$
Пр.22) $\int {\frac{{dx}}{{\sqrt {1 - {x^2}} {{\arcsin }^4}x}}}= $
$\arcsin x = t$
$\frac{1}{{\sqrt {1 - {x^2}} }}dx = dt$
$ = \int {\frac{{dt}}{{{t^4}}}} = \int {{t^{ - 4}}} dt = \frac{{{t^{ - 3}}}}{{ - 3}} + C = \frac{1}{{ - 3{{\arcsin }^3}x}} + C$
Пр.23) $\int {\frac{{x - arctgx}}{{1 + {x^2}}}} dx=$
$\operatorname{arc} tgx = t \Rightarrow x = tgt$
$\frac{1}{{1 + {x^2}}}dx = dt$
$ = \int {\left( {tgt - t} \right)} dt = \int {tgt} dt - \int t dt = \int {\frac{{\sin t}}{{\cos t}}} dt - \frac{{{t^2}}}{2} + C = $
$\cos t = s$
$ - \operatorname{sintdt} = ds$
$\operatorname{sintdt} = - ds$
$ = \int {\frac{{ - ds}}{s}} - \frac{{{t^2}}}{2} + C = - \ln \left| s \right| - \frac{{{t^2}}}{2} + C = - \ln \left| {\cos t} \right| - \frac{{{t^2}}}{2} + C = $
$ = - \ln \left| {\cos \left( {\operatorname{arc} tgx} \right)} \right| - \frac{{\operatorname{arc} t{g^2}x}}{2} + C$
Пр.24) $\int {\frac{{\ln 3x}}{{x\ln 9x}}} dx = \int {\frac{{\ln 3x}}{{x\left( {\ln 3 + \ln 3x} \right)}}} dx = $
$\ln 3x = t$
$\frac{1}{{3x}} \cdot \left( {3x} \right)dx = dt$
$\frac{1}{{3x}}3dx = dt$
$ = \int {\frac{t}{{\ln 3 + t}}} dt = \int {\frac{{t + \ln 3 - \ln 3}}{{\ln 3 + t}}} dt = \int {\frac{{t + \ln 3}}{{\ln 3 + t}}} dt - \int {\frac{{\ln 3}}{{\ln 3 + t}}} dt = $
$ = t - \ln 3\int {\frac{1}{{\ln 3 + t}}} dt = $
$\ln 3 + t = s$
$dt = ds$
$ = t - \ln 3\int {\frac{{ds}}{3}} = t - \ln 3 \cdot \ln \left| s \right| + C = t - \ln 3 \cdot \ln \left| {\ln 3 + t} \right| + C = $
$ = \ln 3x - \ln 3 \cdot \ln \left| {\ln 3 + \ln 3x} \right| + C$