Текст задатака објашњених у видео лекцији:

Пр.6)   Решити $\int {{x^3}\ln xdx} $

Пр.7)   $\int {xarctgdx} $

Пр.8)   $\int {\frac{{\operatorname{arcsinx} dx}}{{{{\sqrt {1 - {x^2}} }^3}}}} $

Пр.6)   $\int {{x^3}\ln xdx} $

\[\begin{array}{*{20}{c}}
\begin{gathered}
u = \ln x \hfill \\
du = \frac{1}{x}dx \hfill \\ 
\end{gathered} &{}&\begin{gathered}
dv = {x^3}dx \hfill \\
v = \int {{x^3}dx = \frac{{{x^4}}}{4} + C} \hfill \\ 
\end{gathered} 
\end{array}\]

$ = \frac{{{x^4}}}{4}\ln x - \int {\frac{{{x^4}}}{4}}  \cdot \frac{1}{x}dx = \frac{{{x^4}}}{4}\ln x - \frac{1}{4}\int {{x^3}} dx = \frac{{{x^4}}}{4}\ln x - \frac{1}{4} \cdot \frac{{{x^4}}}{4} + C = $

$ = \frac{{{x^4}}}{4}\left( {\ln x - \frac{1}{4}} \right) + C$

Пр.7)   $\int {xarctgxdx}  = $

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

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

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

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

Пр.8)   $\int {\frac{{\operatorname{arcsinx} dx}}{{{{\sqrt {1 - {x^2}} }^3}}}}  = \int {\frac{{\operatorname{arcsinx} dx}}{{\left( {1 - {x^2}} \right)\sqrt {1 - {x^2}} }}}  =  $

\[\begin{array}{*{20}{c}}
\begin{gathered}
\operatorname{arcsinx} = t \hfill \\
\frac{1}{{\sqrt {1 - {x^2}} }}dx = dt \hfill \\
\end{gathered} & \Rightarrow &\begin{gathered}
x = \sin t \hfill \\
1 - {x^2} = 1 - {\sin ^2}t = {\cos ^2}t \hfill \\
\end{gathered}
\end{array}\]

$ = \int {\frac{{tdt}}{{{{\cos }^2}t}}}  = $

\[\begin{array}{*{20}{c}}
\begin{gathered}
u = t \hfill \\
du = dt \hfill \\
\end{gathered} &{}&\begin{gathered}
dv = \frac{{dt}}{{{{\cos }^2}t}} \hfill \\
v = \int {\frac{{dt}}{{{{\cos }^2}t}} = tgt + C} \hfill \\
\end{gathered}
\end{array}\]

$ = t \cdot tgt - \int {tgtdt = } t \cdot tgt - \int {\frac{{\sin t}}{{\cos t}}dt = } $

$\cos t = s$

$ - \sin tdt = ds$

$\sin tdt =  - ds$

$ = t \cdot tgt + \int {\frac{{ds}}{s} = } t \cdot tgt + \ln \left| s \right| + C = t \cdot tgt + \ln \left| {\cos t} \right| + C = t \cdot tgt + \ln \left| {\cos t} \right| + C = $

$ = \operatorname{arcsinx}  \cdot tg\left( {\operatorname{arcsinx} } \right) + \ln \left| {\cos \left( {\operatorname{arcsinx} } \right)} \right| + C$