Question

Интегралы
вычислить интегралы​

Answer 1

$\int \frac{(x+4)\, dx}{(x^2+2x+5)(x^2+1)\, x}=Q\\\\\frac{x+4}{(x^2+2x+5)(x^2+1)\, x}=\frac{A}{x}+\frac{Bx+C}{x^2+1}+\frac{Mx+N}{x^2+2x+5}\\\\x+4=A(x^2+1)(x^2+2x+5)+(Bx+C)\cdot x\cdot (x^2+2x+5)+(Mx+N)\cdot x\cdot (x^2+1)\\\\x=0\; \; \to \; \; A=\frac{4}{5\cdot 1}=\frac{4}{5}\\\\x^4\; |\; A+B+M=0\\\\x^3\; |\; 2A+2B+C+N=0\\\\x^2\; |\; 6A+5B+2C+M=0\\\\x^1\; |\; 2A+5C+N=1\\\\x^0\; |\; 5A=4\\\\B=-\frac{9}{10}\; ,\; \; C=-\frac{1}{5}\; ,\; \; M=\frac{1}{10}\; ,\; \; N=\frac{2}{5}$

$Q=\frac{4}{5}\int \frac{dx}{x}-\int \frac{\frac{9}{10}x+\frac{1}{5}}{x^2+1}\, dx+\int \frac{\frac{1}{10}x+\frac{2}{5}}{x^2+2x+5}\, dx=\frac{4}{5}ln|x|-\frac{9}{10\cdot 2}\int \frac{2x\, dx}{x^2+1}+\\\\+\frac{1}{5}\int \frac{dx}{x^2+1}+\frac{1}{10}\int \frac{(x+4)\, dx}{(x+1)^2+4}=\frac{4}{5}\, ln|x|-0,45\cdot ln|x^2+1|+0,2\cdot arctgx+\\\\+0,1\cdot \Big (\frac{1}{2}\int \frac{2(x+1)\, dx}{(x+1)^2+4}+3\int \frac{dx}{(x+1)^2+4}\Big )=\frac{4}{5}\, ln|x|-0,45\cdot ln(x^2+1)+0,2\cdot arctgx+$

$+0,05\cdot ln|(x+1)^2+4|+0,3\cdot \frac{1}{2}\, arctg\frac{x+1}{2}+C=\\\\=\frac{4}{5}\, ln|x|-0,45\cdot ln(x^2+1)+0,2\cdot arctgx+0,05\cdot ln(x^2+2x+5)+\\\\+0,15\cdot arctg\frac{x+1}{2}+C$