역함수를 이용한 치환적분

\(\int f(x)\,dx=xf(x)-\int xf'(x)\,dx+xf(x)-\int f^{-1}(f(x))f'(x)\,dx+xf(x)-G(f(x))\)

여기서 \(G(x)= \int f^{-1}(x)\,dx\)

 

문제  

\(\int \sqrt{\frac{x}{1-x}}\,dx\)

\(G(x)=\int f^{-1}(x)\,dx= \int\frac{x^2}{1+x^2}\,dx=\int(1-\frac{1}{1+x^2})\,dx=x-\arctan x+C\)

따라서, 

\(\sqrt{\frac{x}{1-x}}\,dx=(x-1)\sqrt{\frac{x}{1-x}}+\arctan{\sqrt{\frac{x}{1-x}}}+C\)

\(\int \sqrt{\frac{x}{1-x}}\,dx= (x-1)\sqrt{\frac{x}{1-x}}+\arctan{\sqrt{\frac{x}{1-x}}}+C\)