Неодређени интеграли – парцијална интеграција 3

Задаци

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)   $\int {{x^3}\ln xdx}$

$$\begin{array}{cccccccccccccccccccc} \begin{aligned}&u = \ln x \\ &du = \frac{1}{x}dx\end{aligned} &{}&\begin{aligned}&dv = {x^3}dx \\ &v = \int {{x^3}dx = \frac{{{x^4}}}{4} + C}\end{aligned} \end{array}$$

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

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

Пр.7)   $\int {xarctgxdx} =$

$$\begin{array}{cccccccccccccccccccc} \begin{aligned}&u = arctgx \\ &du = \frac{1}{{1 + {x^2}}}dx\end{aligned} &{}&\begin{aligned}&dv = xdx \\ &v = \int {xdx = \frac{{{x^2}}}{2} + C}\end{aligned} \end{array}$$

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

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

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

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

$$\begin{array}{cccccccccccccccccccc} \begin{aligned}&\operatorname{arcsinx} = t \\ &\frac{1}{{\sqrt {1 - {x^2}} }}dx = dt\end{aligned} & \Rightarrow &\begin{aligned}&x = \sin t \\ &1 - {x^2} = 1 - {\sin ^2}t = {\cos ^2}t\end{aligned} \end{array}$$

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

$$\begin{array}{cccccccccccccccccccc} \begin{aligned}&u = t \\ &du = dt\end{aligned} &{}&\begin{aligned}&dv = \frac{{dt}}{{{{\cos }^2}t}} \\ &v = \int {\frac{{dt}}{{{{\cos }^2}t}} = tgt + C}\end{aligned} \end{array}$$

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

$\cos t = s$

$- \sin tdt = ds$

$\sin tdt = - ds$

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

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