1. integral (x+1)*cos3xdx 2. integral (x^2+3x+2)ln xdx 3. integral(4x^2+6x-7)ln xdx

Question 1

1. integral (x+1)*cos3xdx
2. integral (x^2+3x+2)ln xdx
3. integral(4x^2+6x-7)ln xdx

Question 2

Решите задачу:

$\int(x+1)*cos3x\ dx\\u=x+1\ \ \ \ \ \ \ \ \ du=dx\\dv=cos3xdx\ \ \ v=\int cos3xdx=\frac{1}{3}sin3x\\\int(x+1)*cos3x\ dx=\frac{(x+1)sin3x}{3}-\frac{1}{3}\int sin3xdx=\\=\frac{(x+1)sin3x}{3}+\frac{cos3x}{9}+C\\\\(\frac{(x+1)sin3x}{3}+\frac{cos3x}{9}+C)`=\frac{sin3x+3cos3x(x+1)}{3}-\frac{27sin3x}{81}=\\=\frac{sin3x}{3}+cos3x(x+1)-\frac{sin3x}{3}=cos3x(x+1)$

$\int(x^2+3x+2)ln x\ dx\\u=lnx\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ du=\frac{dx}{x}\\dv=(x^2+3x+2)dx\ \ \ v=\int(x^2+3x+2)dx=\frac{x^3}{3}+\frac{3x^2}{2}+2x\\\\\int(x^2+3x+2)ln x\ dx=(\frac{x^3}{3}+\frac{3x^2}{2}+2x)lnx-\int\frac{(\frac{x^3}{3}+\frac{3x^2}{2}+2x)dx}{x}=\\=(\frac{x^3}{3}+\frac{3x^2}{2}+2x)lnx-\int(\frac{x^2}{3}+\frac{3x}{2}+2)dx=\\=(\frac{x^3}{3}+\frac{3x^2}{2}+2x)lnx-\frac{x^3}{9}-\frac{3x^2}{4}-2x+C$
$((\frac{x^3}{3}+\frac{3x^2}{2}+2x)lnx-\frac{x^3}{9}-\frac{3x^2}{4}-2x+C)`=(x^2+3x+2)lnx+\frac{x^2}{3}+\\\\+\frac{3x}{2}+2-\frac{x^2}{3}-\frac{3x}{2}-2=(x^2+3x+2)lnx$

$\int(4x^2+6x-7)ln x\ dx\\u=lnx\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ du=\frac{dx}{x}\\dv=(4x^2+6x-7)dx\ \ \ \ \ v=\int(4x^2+6x-7)dx=\frac{4x^3}{3}+3x^2-7x\\\\\int(4x^2+6x-7)ln x\ dx=(\frac{4x^3}{3}+3x^2-7x)lnx-\int\frac{(\frac{4x^3}{3}+3x^2-7x)dx}{x}=\\=(\frac{4x^3}{3}+3x^2-7x)lnx-\int(\frac{4x^2}{3}+3x-7)dx=\\=(\frac{4x^3}{3}+3x^2-7x)lnx-\frac{4x^3}{9}-\frac{3x^2}{2}+7x+C$
$((\frac{4x^3}{3}+3x^2-7x)lnx-\frac{4x^3}{9}-\frac{3x^2}{2}+7x+C)`=(4x^2+6x-7)lnx+\\\\+\frac{4x^2}{3}+3x-7-\frac{4x^2}{3}-3x+7=(4x^2+6x-7)lnx$