Неопределени интеграли - 2

[Parasyte]

Не държа да се решат всички, а част от тях, за да взаимствам от решението и да реша останалите.

[Nathi123]

27.1. зад. Два пъти интегриране по части: [tex]I= \int x^{2}cos3xdx= \frac{1}{3} \int x^{2}dsin3x= \frac{x^{2}.sin3x}{3}- \frac{1}{3}\int xsin3x =\frac{x^{2}.sin3x}{3}+\frac{2}{9}\int xdcos3x\Rightarrow[/tex]
[tex]I= \frac{x^{2}.sin3x}{3}+\frac{2x.cos3x}{9}-\frac{2}{9}\int cos3x dx= \frac{x^{2}.sin3x}{3} +\frac{2x.cos3x}{9}-\frac{2}{27}sin3x +C.[/tex]

[Nathi123]

28.1 I=[tex]\int xlnxdx=\frac{1}{2}\int lnxdx^{2} = \frac{x^{2}.lnx}{2} -\frac{1}{2}\int \frac{x^{2}dx}{x}= \frac{x^{2}.lnx}{2} - \frac{1}{2}\int xdx \Rightarrow[/tex]
[tex]I= \frac{x^{2}.lnx}{2} -\frac{1}{4}x^{2}+C.[/tex]

[aifC]

29.1 [tex]\int x^{3}ln(x-1)dx = \frac{ln(x-1)x^{4}}{4} - \int \frac{x^{4}}{4(x-1)}dx; I = \int \frac{x^{4}}{4(x-1)}dx = \frac{1}{4} \int \frac{x^{4}}{x-1}dx ;[/tex]
[tex]u = x-1 \Rightarrow x^{4} = (u+1)^{4} \Rightarrow \int\frac{(u+1)^{4}}{u}du = \int(u^{3}+4u^{2}+6u+\frac{1}{u}+4)du = \int u^{3}du + 4\int u^{2}du + 6\int udu + \int\frac{1}{u}du + 4\int 1du = \frac{ln(x-1)x^{4}}{4} -x-\frac{(x-1)^{4}}{16} - \frac{(x-1)^{3}}{3} -\frac{3(x-1)^{2}}{4} - \frac{ln(x-1)}{4} +1 + C;[/tex]

[aifC]

30.1 [tex]\int x^{2}ln^{2}(x)dx = \frac{x^{3}ln^{2}(x)}{3} - \int \frac{2x^{2}ln(x)}{3}dx \Rightarrow \frac{2}{3}\int x^{2}ln(x)dx = \frac{x^{3}ln(x)}{3} - \int \frac{x^{2}}{3}dx = \frac{1}{3} \int x^{2}dx = \frac{x^{3}}{3} \Rightarrow \frac{x^{3}ln^{2}(x)}{3} - \frac{2x^{3}ln(x)}{9} + \frac{2x^{3}}{27 } = \frac{x^{3}(9ln^{2}(x)-6ln(x)+2}{27} +C;[/tex]

[aifC]

31.1 [tex]\int ln(2x-1)dx; u = 2x-1 \Rightarrow dx = \frac{1}{2}du \Rightarrow \frac{1}{2} \int ln(u)du = uln(u)-\int 1du = \frac{(2x-1)(ln(2x-1)-1}{2}+C[/tex]

[aifC]

32.1 [tex]\int arctg(x)dx = xarctg(x)-\int \frac{x}{x^{2}+1}; u = x^{2}+1 \Rightarrow dx = \frac{1}{2x}du; \frac{1}{2}\int \frac{1}{u}du = xarctg(x) - \frac{ln(x^{2}+1)}{2} + C;[/tex]

[aifC]

33.1 [tex]\int xarccotg dx = \frac{x^{2}arccotg(x)}{2} - \int -\frac{x^{2}}{2(x^{2}+1)}dx = -\frac{1}{2}\int \frac{x^{2}}{x^{2}+1}dx = \int \left(1 - \frac{1}{x^{2}+1} \right) = \int 1dx - \int \frac{1}{x^{2}+1}dx = \frac{x-arctg(x)}{2}+\frac{x^{2}arccotg(x)}{2} + C;[/tex]

[aifC]

34.1 [tex]\int \frac{2}{\sqrt{3-2x-x^{2}}} = 2 \int \frac{1}{\sqrt{3-2x-x^{2}}} = \int \frac{1}{\sqrt{4-(x+1)^{2}}} ; u = \frac{x+1}{2}, dx = 2du \Rightarrow \int \frac{1}{\sqrt{1-u^{2}}} = arcsin(u) \Rightarrow -2arcsin \left(\frac{-2x-2}{4} \right)+C;[/tex]