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$
