Question

решите интегралы, пожалст

Answer 1

$1)\ \ \int \dfrac{26x^2-66x-18}{x^3-9x^2+36x-54}\, dx=\int \dfrac{26x^2-66x-18}{(x-3)(x^2-6x+18)}=I\\\\\\\dfrac{26x^2-66x-18}{(x-3)(x^2-6x+18)}=\dfrac{A}{x-3}+\dfrac{Bx+C}{x^2-6x+18}\\\\\\26x^2-66x-18=A(x^2-6x+18)+(Bx+C)(x-3)\\\\x=3\ \ \to \ \ A=\dfrac{26x^2-66x-18}{x^2-6x+18}\ \Big|_{x=3}=2\\\\x^2\ |\ A+B=26\ \ \ \ \ \ ,\ \ \ \ \ \ \ \ \ B=26-A=25-2=24\\x^1\ |\ -6A-3B+C=-66\\x^0\ |\ 18A-3C=-18\ \ \ \ ,\ \ \ \ \ \ \ \ C=6A+6=12+6=18$

$I=\int \dfrac{2\, dx}{x-3}+\int \dfrac{24x+18}{x^2-6x+18}=2\, ln|x-3|+\int \dfrac{24x+18}{(x-3)^2+9}\, dx\ ;\\\\\\\\\star \ \int \dfrac{24x+18}{(x-3)^2+9}\, dx=\Big[\ t=x-3\ ,\ x=t+3\ ,\ dx=dt\ \Big]=\int \dfrac{24t+90}{t^2+9}\, dt=\\\\\\=12\int \dfrac{d(t^2+9)}{t^2+9}+90\int \dfrac{dt}{t^2+9}=12\cdot ln|t^2+9|+90\cdot \dfrac{1}{3}\cdot arctg\dfrac{t}{3}+C_1=\\\\\\=12\cdot ln|\, x^2-6x+18\, |+30\cdot arctg\dfrac{x-3}{3}+C_1\ ;\\\\\\I=2\, ln|x-3|+12\cdot ln|\, x^2-6x+18\, |+30\cdot arctg\dfrac{x-3}{3}+C\ .$

$2)\ \ \int (4x+4)\, cos(3x+18)\, dx=\Big[\ u=4x+4\ ,\ du=4\, dx\ ,v=cos(3x+18)\ ,\\\\\\dv=\dfrac{1}{3}\, sin(3x+8)\ \Big]=\dfrac{1}{3}\cdot (4x+4)\, sin(3x+8)-\dfrac{4}{3}\int sin(3x+8)\, dx=\\\\\\=\dfrac{1}{3}\cdot (4x+4)\, sin(3x+8)+\dfrac{4}{9}\, cos(3x+8)+C\ .$