Question

Помогите, пожалуйста)

Answer 1

$int frac{1}{x}sqrt{frac{1-x}{1+x}}dx=[t^2=frac{1-x}{1+x},; t^2(1+x)=1-x,; x(t^2+1)=1-t^2,; \\x=frac{1-t^2}{t^2+1}=-1+frac{2}{t^2+1},; dx=frac{-4t}{(t^2+1)^2}dt, ]=\\=int frac{t^2+1}{1-t^2}cdot frac{t(-4t)}{(t^2+1)^2}dt=4int frac{t^2}{(t^2-1)(t^2+1)}dt=4int frac{t^2}{(t-1)(t+1)(t^2+1)}dt=I\\frac{t^2}{(t-1)(t+1)(t^2+1)}=frac{A}{t-1}+frac{B}{t+1}+frac{Ct+D}{t^2+1}; Rightarrow \\t^2=(At+A)(t^2+1)+(Bt-B)(t^2+1)+(Ct+D)(t^2-1)\\t=1,; A=frac{1}{4};\\ t=-1,; B=-frac{1}{4}$

$t^1, |, A+B-C=0,; frac{1}{4}-frac{1}{4}-C=0,; C=0\\t^0, |; A-B-D=0,; D=frac{1}{2}\\I=4int (frac{1}{4(t-1)}-frac{1}{4(t+1)}+frac{1}{2(t^2+1)})dt=\\=ln|t-1|-ln|t+1|+frac{1}{2}cdot arctgt+C=\\=ln|sqrt{frac{1-x}{1+x}}-1|-ln|sqrt{frac{1-x}{1+x}}+1|+frac{1}{2}cdot arctgsqrt{frac{1-x}{1+x}}+C=\\=ln|frac{sqrt{1-x}-sqrt{1+x}}{sqrt{1-x}+sqrt{1+x}}|+frac{1}{2}cdot arctgsqrt{frac{1-x}{1+x}}+C$