Question

Пожалуйста, РЕШИТЕ это!!!! Очень срочно!

Answer 1

Ответ:

$\displaystyle 1)\ \ \int \frac{dx}{x^2+4x+9}=\int \frac{dx}{(x+2)^2+5}=\int \frac{d(x+2)}{(x+2)^2+5}=\\\\\\=\Big[\ \int \frac{du}{u^2+a^2}=\frac{1}{a}arctg\frac{u}{a}+C\ \Big]=\frac{1}{\sqrt5}\cdot arctg\frac{x+2}{\sqrt5}+C$

$\displaystyle 2)\ \ \int \frac{(x+1)\, dx}{4x^2+4x-3}=\int \frac{(x+1)\, dx}{(2x+1)^2-4}=\Big[\ 2x+1=t\ ,\ x=\frac{t}{2}-\frac{1}{2}\ ,\\\\\\dx=\frac{1}{2}dt\Big]=\frac{1}{2}\int \frac{\dfrac{t}{2}+\dfrac{1}{2}}{t^2-4}\, dt=\frac{1}{4}\cdot \frac{1}{2}\int \frac{2t\, dt}{t^2-4}+\frac{1}{4}\int \frac{dt}{t^2-4}=\\\\\\=\frac{1}{8}\cdot ln|t^2-4|+\frac{1}{4}\cdot \frac{1}{4}\, ln\Big|\frac{t-2}{t+2}\Big|+lnC=\\\\\\=\frac{1}{8}\cdot ln\Big|4x^2+4x-3\Big|+\frac{1}{16}\, ln\Big|\frac{2x-1}{2x+3}\Big|+lnC$

$\displaystyle 3)\ \ \int \frac{(2x+3)\, dx}{x^2+3x-10}=\int\frac{d(x^2+3x-10)}{x^2+3x-10}=ln\Big|x^2+3x-10\Big|+C\\\\\\\\\star \ \ \int \frac{du}{u}=ln|u|+C\ \ \star$