Интегриране по части

[Гост]

Помощ

[aifC]

25.3 [tex]\int xe^{x}dx = xe^{x} - \int e^{x}dx = (x-1)e^{x}+C;[/tex]

26.3 [tex]\int xcos(3x)dx = \frac{xsin(3x)}{3} - \int \frac{sin(3x)}{3}dx; u=3x \rightarrow dx = \frac{1}{3}du \Rightarrow \frac{1}{9} \int sin(u)du = -cos(u) \Rightarrow \frac{3xsin(3x)+cos(3x)}{9} + C;[/tex]

27.3 [tex]\int x^{2}sin(3-x)dx = -\int-2cos(x-3)xdx-cos(x-3)x^{2} = -2\int cos(x-3)xdx = -2sin(x-3)x - \int sin(x-3)dx = \int sin(u)du = -cos(u) \Rightarrow
x(cos(x-3)x-2sin(x-3))-2cos(x-3) +C;[/tex]

28.3 [tex]\int x^{3}ln(x)dx = \frac{x^{4}ln(x)}{4} - \int \frac{x^{3}}{4}dx = \frac{1}{4} \int x^{3}dx = \frac{x^{4}}{4} = \frac{x^{4}(4ln(x)-1)}{16}+C;[/tex]

[aifC]

29.3 [tex]\int x^{2}ln(x+1)dx = \frac{x^{3}ln(x+1)}{3} - \int \frac{x^{3}}{3(x+1)}dx = \frac{1}{3} \int \frac{x^{3}}{x+1}dx; u=x+1 \rightarrow dx = du;
\int \frac{(u-1)^{3}}{u}du = \int \left(u^{2}-3u-\frac{1}{u}+3 \right)du = \int u^{2}du - 3\int udu - \int \frac{1}{u}du +3\int 1du =
\frac{(6x^{3}+6)ln(x+1)-2x^{3}+3x^{2}-6x}{18} +C;[/tex]

30.3 [tex]\int xln^{2}xdx = \frac{x^{2}ln^{2}(x)}{2} - \int xln(x)dx = \frac{x^{2}ln(x)}{2} - \int \frac{x}{2}dx = \frac{1}{2} \int xdx = \frac{x^{2}}{2} \Rightarrow \frac{x^{2}(2ln^{2}(x)-2ln(x)+1}{4}+C;[/tex]

31.3 [tex]\int ln(2-3x)dx; u=2-3x \rightarrow dx= -\frac{1}{3}du \Rightarrow -\frac{1}{3} \int ln(u)du = uln(u) - \int 1du = \frac{(ln(2-3x)-1)(3x-2)}{3}+C;[/tex]