Question

$\int \sqrt{\frac{1-x}{x}}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{\sqrt{1-x}}{\sqrt{x}}dx \\ =\int \left(\frac{\sqrt{1-x}·\sqrt{1-x}}{\sqrt{x}·\sqrt{1-x}}\right)dx \\ =\int \frac{\left(1-x\right)}{\sqrt{x-x^2}}dx \\ \mathrm{Let}\left(1-x\right)=\mathit{}A\frac{d}{dx}\left(x-x^2\right)+B \\ \Rightarrow 1-x=A\left(1-2x\right)+B \\ \Rightarrow 1-x=-\left(2A\right)x+A+B \\ \mathrm{Equating}\mathrm{coefficients}\mathrm{of}\mathrm{like}\mathrm{terms} \\ -2A=-1 \\ \Rightarrow A=\frac{1}{2} \\ \&\mathit{}A+B=1 \\ \Rightarrow \frac{1}{2}+B=1 \\ \therefore B=\frac{1}{2} \\ \therefore I=\int \frac{{\displaystyle \frac{1}{2}}\left(1-2x\right)+{\displaystyle \frac{1}{2}}}{\sqrt{x-x^2}}dx \\ =\frac{1}{2}\int \frac{\left(1-2x\right)}{\sqrt{x-x^2}}dx+\frac{1}{2}\int \frac{1}{\sqrt{x-x^2+{\left({\displaystyle \frac{1}{2}}\right)}^2-{\left({\displaystyle \frac{1}{2}}\right)}^2}}dx \\ =\frac{1}{2}\int \frac{\left(1-2x\right)}{\sqrt{x-x^2}}dx+\frac{1}{2}\int \frac{1}{\sqrt{{\left({\displaystyle \frac{1}{2}}\right)}^2-\left(x^2-x+{\displaystyle \frac{1}{2^2}}\right)}}dx \\ =\frac{1}{2}\int \frac{\left(1-2x\right)}{\sqrt{x-x^2}}dx+\frac{1}{2}\int \frac{1}{\sqrt{{\left({\displaystyle \frac{1}{2}}\right)}^2-{\left(x-{\displaystyle \frac{1}{2}}\right)}^2}}dx \end{gathered}$$

$$\begin{gathered} \mathrm{Putting}x-x^2=t\mathrm{in}\mathrm{the}\mathrm{first}\mathrm{integral} \\ \Rightarrow \left(1-2x\right)dx=dt \\ \therefore I=\frac{1}{2}\int \frac{1}{\sqrt{t}}dt+\frac{1}{2}\int \frac{1}{\sqrt{{\left({\displaystyle \frac{1}{2}}\right)}^2-{\left({\displaystyle x-\frac{1}{2}}\right)}^2}}dx \\ =\frac{1}{2}\int t^{-\frac{1}{2}}dt+\frac{1}{2}\int \frac{dx}{\sqrt{{\left({\displaystyle \frac{1}{2}}\right)}^2-{\left({\displaystyle x-\frac{1}{2}}\right)}^2}} \\ =\frac{1}{2}\times 2t^{\frac{1}{2}}+\frac{1}{2}\times {\sin }^{-1}\left(\frac{x-{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{1}{2}}}\right)+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left[\because \int \frac{1}{\sqrt{a^{\mathit{2}}\mathit{-}{x}^{\mathit{2}}}}dx={\sin }^{-1}\frac{x}{a}+C\right] \\ =\sqrt{t}+\frac{1}{2}{\sin }^{-1}\left(2x-1\right)+C \\ =\sqrt{x-x^2}+\frac{1}{2}{\sin }^{-1}\left(2x-1\right)+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left[\mathit{\because }\mathit{}t\mathit{=}\mathit{}x\mathit{-}{x}^{\mathit{2}}\right] \end{gathered}$$