Question

Неопределенный интеграл

Answer 1

$\int \Big (\frac{1-x}{1+x}\Big )^{3/2}dx=\left \int \sqrt{ (\frac{1-x}{1+x} )^3}dx=\Big [\, \frac{1-x}{1+x}=t^2\; ,1-x=t^2(1+x)\; ,$

$1-x=t^2+t^2x\; ,\; 1-t^2=x(1+t^2)\; ,\; x=\frac{1-t^2}{1+t^2}\; ,\\\\dx=\frac{-2t(1+t^2)-2t(1-t^2)}{(1+t^2)^2}dt=\frac{-4t}{(1+t^2)^2}dt\, \Big ]=\int \sqrt{t^6}\cdot \frac{-4t\, dt}{(1+t^2)^2}=\\\\=-4\int \frac{t^3\cdot t\cdot dt}{(1+t^2)^2} =-4\int \frac{t^4\cdot dt}{t^4+2t^2+1}=-4\int \frac{(t^4+2t^2+1)-2t^2-1}{t^4+2t^2+1}\, dt=\\\\=-4\int \Big (1-\frac{2t^2+1}{(t^2+1)^2}\Big )dt=-4\int dt+4\int \frac{(t^2+1)+t^2}{(t^2+1)^2}\, dt=\\\\=-4t+4\int \Big (\frac{1}{t^2+1}+\frac{t^2}{(t^2+1)^2}\Big )dt=$

$=-4t+4\int \frac{dt}{t^2+1}+4\int \frac{t\cdot t\, dt}{(t^2+1)^2}=-4t+4arctgt+4\int t\cdot \frac{t\, dt}{(t^2+1)^2}=I\\\\\\\star \; \; \int t\cdot \frac{t\cdot dt}{(t^2+1)^2}=\Big [\, u=t\; ,\; du=dt,\; dv=\frac{t\cdot dt}{(t^2+1)^2},\; v=\frac{1}{2}\int \frac{2t\, dt}{(t^2+1)^2}=\\\\=\frac{1}{2}\int \frac{d(t^2+1)}{(t^2+1)^2} =\frac{1}{2}\cdot \frac{(t^2+1)^{-1}}{-1}=-\frac{1}{2(t^2+1)}\, \Big ]=uv-\int v\cdot du=\\\\=-\frac{t}{2(t^2+1)}+\frac{1}{2}\int \frac{dt}{t^2+1}=-\frac{t}{2(t^2+1)}+\frac{1}{2}arctgt+C_1\, ;$


$I=-4t+4arctgt+4\cdot \Big (-\frac{t}{2(t^2+1)}+\frac{1}{2}arctgt\Big )+C=\\\\=-4t+4arctgt-\frac{2t}{t^2+1}+2arctgt+C=\\\\=-4 \sqrt{\frac{1-x}{1+x}} +6arctg\sqrt{\frac{1-x}{1+x}}-\frac{2\sqrt{1-x}}{\sqrt{1+x}\cdot (\frac{1-x}{1+x}+1)}+C=\\\\=-4 \sqrt{\frac{1-x}{1+x}}+6arctg\sqrt{\frac{1-x}{1+x}}-\sqrt{1-x^2}+C$