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Неодређени интеграли – парцијална интеграција 1


Задаци


Текст задатака објашњених у видео лекцији:

Пр.1)   Решити $\int {\ln xdx} $

Пр.2)   $\int {arctgxdx} $

Пр.3)   $\int {\ln \left( {{x^2} + 1} \right)} dx$


 

Пр.1)   $\int {\ln xdx} =$

\[\begin{array}{*{20}{c}}
\begin{gathered}
\ln x = u \hfill \\
\frac{1}{x}dx = du \hfill \\
\end{gathered} &{}&\begin{gathered}
dv = dx \hfill \\
\int {dv} = \int {dx} \hfill \\
v = x + C \hfill \\
\end{gathered}
\end{array}\]

$= \ln x \cdot x - \int {x \cdot \frac{1}{x}} dx = x\ln x - x + C = x\left( {\ln x - 1} \right) + C$

 

Пр.2)   $\int {arctgxdx}= $

\[\begin{array}{*{20}{c}}
\begin{gathered}
arctgx = u \hfill \\
\frac{1}{{1 + {x^2}}}dx = du \hfill \\
\end{gathered} &{}&\begin{gathered}
dv = dx/\smallint \hfill \\
\int {dv = \int {dx} } \hfill \\
v = x + C \hfill \\
\end{gathered}
\end{array}\]

$ = x \cdot arctgx - \int {\frac{{xdx}}{{1 + {x^2}}}}  = $

Смена:  

$1 + {x^2} = t/'$

$2xdx = dt$

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

$ = x \cdot arctgx - \int {\frac{{\frac{{dt}}{2}}}{t}}  = x \cdot arctgx - \frac{1}{2}\ln \left| t \right| + C = x \cdot arctgx - \frac{1}{2}\ln \left| {1 + {x^2}} \right| + C$

 

Пр.3)   $\int {\ln \left( {{x^2} + 1} \right)} dx=$

\[\begin{array}{*{20}{c}}
\begin{gathered}
u = \ln \left( {{x^2} + 1} \right) \hfill \\
du = \frac{1}{{{x^2} + 1}}{\left( {{x^2} + 1} \right)^\prime }dx \hfill \\
du = \frac{1}{{{x^2} + 1}}2xdx \hfill \\
\end{gathered} &{}&\begin{gathered}
dv = dx/\smallint \hfill \\
\smallint dv = \smallint dx \hfill \\
v = x + C \hfill \\
\end{gathered}
\end{array}\]

$ = x\ln \left( {{x^2} + 1} \right) - \int {x \cdot } \frac{1}{{{x^2} + 1}}2xdx = x\ln \left( {{x^2} + 1} \right) - 2\int {\frac{{{x^2} + 1 - 1}}{{{x^2} + 1}}dx}  = $

$ = x\ln \left( {{x^2} + 1} \right) - 2\left( {\int {\frac{{{x^2} + 1}}{{{x^2} + 1}}dx - \int {\frac{{dx}}{{{x^2} + 1}}} } } \right) = x\ln \left( {{x^2} + 1} \right) - 2\left( {x - arctgx} \right) + C$

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