Question

найти интеграл (1/x)*√((x-1)/(x+1))dx (замена t=√((x-1)/(x+1)), t^2*(x+1)=x-1, x= -(t^2+1)/(t^2-1), dx=(-(t^2+1)/(t^2-1))dt).​

Answer 1

$\displaystyle\int\frac{1}{x}\sqrt{\frac{x-1}{x+1}}dx=-\int\frac{t^2-1}{t^2+1}*t*\frac{4tdt}{(t^2-1)^2}=\\=-4\int\frac{t^2}{(t^2+1)(t^2-1)}dt=-4\int(\frac{1}{2(t^2+1)}+\frac{1}{4(t-1)}-\frac{1}{4(t+1)})dt=\\=-2arctgt-ln|t-1|+ln|t+1|+C=\\=-2arctgt+ln|\frac{t+1}{t-1}|+C=-2arctg\sqrt{\frac{x-1}{x+1}}+ln|\frac{\sqrt{\frac{x-1}{x+1}}+1}{\sqrt{\frac{x-1}{x+1}}-1}|+C$

$\displaystyle t=\sqrt{\frac{x-1}{x+1}}\to t^2=\frac{x-1}{x+1}\\(x+1)t^2=x-1\to t^2x+t^2=x-1\\x(t^2-1)=-(t^2+1)\to x=-\frac{t^2+1}{t^2-1}\to dx=\frac{4tdt}{(t^2-1)^2}\\\\\\\frac{t^2}{(t^2+1)(t^2-1)}=\frac{At+B}{t^2+1}+\frac{C}{t-1}+\frac{D}{t+1}=\\=\frac{1}{2(t^2+1)}+\frac{1}{4(t-1)}-\frac{1}{4(t+1)}\\\\\\t^2=A(t^3-t)+B(t^2-1)+C(t^3+t^2+t+1)+D(t^3-t^2+t-1)\\t^3|0=A+C+D\\t^2|1=B+C-D\\t|0=-A+C+D\\t^0|0=-B+C-D\\A=0;B=\frac{1}{2};C=\frac{1}{4};D=-\frac{1}{4}$