Question

Интегрирование иррациональностей.

Answer 1

$\int \frac{x+\sqrt{1+x+x^2}}{1+x+\sqrt{1+x+x^2}}dx=I\\\\\star \; \; \sqrt{1+x+x^2}=x+t\; ;\; \; 1+x+x^2=(x+t)^2\; ;\\\\1+x+x^2=x^2+2xt+t^2\; ;\; \; x-2xt=t^2-1\; ;\; \; x(1-2t)=t^2-1\; ;\\\\x=\frac{t^2-1}{1-2t}\; ;\; \; \sqrt{1+x+x^2}=\frac{t^2-1}{1-2t} +t=\frac{t^2-1+t-2t^2}{1-2t}=\frac{-t^2+t-1}{1-2t}\; ;\\\\dx=\frac{2t(1-2t)+2(t^2-1)}{(1-2t)^2}dt=\frac{-2t^2+2t-2}{(1-2t)^2} dt\; \; \star \\\\I=\int \frac{\frac{t^2-1}{1-2t}+\frac{-t^2+t-1}{1-2t}}{1+\frac{t^2-1}{1-2t}+\frac{-t^2+t-1}{1-2t}}\cdot \frac{-2t^2+2t-2}{(1-2t)^2}dt=$

$=\int \frac{(t-2)\cdot (-2)\cdot (t^2-t+1)}{-(t+2)(1-2t)^2}dt=2\int \frac{t^3-t^2+t-2t^2+2t-2}{t-4t^2+4t^3+2-8t+16t^2}dt=\\\\=2\int \frac{t^3-3t^2+3t-2}{4t^3+12t^2-7t+2}dt=2\int \Big (\frac{1}{4} +\frac{-6t^2+\frac{19}{4}t-\frac{5}{2}}{(t+2)(1-2t)^2}\Big )dt=\\\\=\frac{1}{2}t-\frac{1}{2}\int \frac{24t^2-19t+10}{(t+2)(1-2t)^2}dt\; ;\\\\\frac{24t^2-19t+10}{(t+2)(1-2t)^2}=\frac{A}{t+2}+\frac{B}{(1-2t)^2}+\frac{C}{1-2t}\\\\24t^2-19t+10=A(1-2t)^2+B(t+2)+C(t+2)(1-2t)\\\\t^2|\; 24=4A-2C\\t\; |\; -19=-4A+B-3C\\t^0|\; 10=A+2B+2C$

$A=\frac{144}{25}\; ,\; \; B=5\; ,\; \; C=-\frac{12}{25}\\\\I=\frac{1}{2}t-\frac{1}{2}\int \Big (\frac{144/25}{t+2}+\frac{5}{(1-2t)^2}-\frac{12/25}{1-2t}\Big )dt =\\\\=\frac{1}{2}t-\frac{72}{25}\cdot ln|t+2|-\frac{5}{2}\cdot \frac{-1}{-2(1-2t)}-\frac{6}{25}\cdot \frac{-1}{2}\cdot ln|1-2t|+C=\\\\=\frac{1}{2}t-\frac{72}{25}\cdot ln|t+2|-\frac{5}{4(1-2t)}+\frac{3}{25}\cdot ln|1-2t|+C\; ,\; t=\sqrt{1+x+x^2}-x\; .$