Question

Решить интегралы 1 и 2 примеры, надо.

Answer 1

$1)\; \; \int \frac{sinx\, dx}{1+3cosx}=\Big [\; u=1+3cosx\; ,\; dt=-3sinx\, dx\; \Big ]=-\frac{1}{3}\int \frac{dt}{t}=\\\\=-\frac{1}{3}\cdot ln|t|+C=-\frac{1}{3}\cdot ln|1+3cosx|+C\; ;$

$2)\; \; \int \frac{dx}{x(x^2+1)}=Q\\\\\frac{1}{x(x^2+1)}=\frac{A}{x}+\frac{Bx+C}{x^2+1}=\frac{A(x^2+1)+x(Bx+C)}{x(x^2+1)}\\\\1=A(x^2+1)+x(Bx+C)\\\\x=0:\; \; 1=A\\\\x^2\; |\; A+B=0\; \; \to \; \; 1+B=0\; ,\; B=-1\\\\x^1\; |\; C=0\\\\Q=\int \Big (\frac{1}{x}-\frac{x}{x^2+1}\Big )dx=\int \frac{dx}{x}-\frac{1}{2}\int \frac{2x\, dx}{x^2+1}=ln|x|-\frac{1}{2}\int \frac{d(x^2+1)}{x^2+1}=\\\\=\Big [\; \int \frac{du}{u}=ln|u|+C\; \Big ]=ln|x|-\frac{1}{2}\, ln|x^2+1|+C=ln\frac{|x|}{\sqrt{x^2+1}}+C\; .$