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

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

$displaystyle t=sqrt{frac{x-1}{x+1}}to t^2=frac{x-1}{x+1}\(x+1)t^2=x-1to 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}$