Question:

Put $x = t^2 \Rightarrow dx = 2t dt$<br>$$I = \int \sqrt{\frac{1-t}{1+t}} \times \frac{2t dt}{t^2} = \int \frac{2(1-t) dt}{t\sqrt{1-t^2}} = 2\int \frac{dt}{t\sqrt{1-t^2}} - 2\int \frac{dt}{\sqrt{1-t^2}} = 2\int \frac{dt}{t^2\sqrt{\frac{1}{t^2}-1}} - 2\sin^{-1}(t) + C$$<br>$$= 1\int \frac{-du}{\sqrt{u^2-1}} - 2\sin^{-1}(t) + C \left[ \text{Put } \frac{1}{t} = u \Rightarrow \frac{dt}{t^2} = -du \right]$$<br>$$= -2\ln|u + \sqrt{u^2-1}| - 2\sin^{-1} t + C = -2\ln\left|\frac{1}{\sqrt{x}} + \sqrt{\frac{1}{x}-1}\right| - 2\sin^{-1}\sqrt{x} + C$$