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

Question

$$\int\frac{\sqrt{1-x}}{x}\,dx$$ $$\int \:uv'=uv-\int \:u'v$$ $$u=\frac{\sqrt{1-x}}{x},\:\:u'=\frac{x-2}{2\sqrt{1-x}x^2},\:\:v'=1,\:\:v=x$$

Answer 1

HINT:

Let $\sqrt{1-x}=y\implies1-x=y^2\iff x=1-y^2 \implies-dx=2y\ dy$

$$\int\frac{\sqrt{1-x}}xdx=-\int\frac y{1-y^2}2y\ dy$$

But $\dfrac{-2y^2}{1-y^2}=\dfrac{2-2y^2-2}{1-y^2}=2-2\cdot\dfrac1{1-y^2}$

Answer 2

setting $t=\sqrt{1-x}$ we get $x=1-t^2$ and $dx=-2tdt$ and our integral is given by $$\int\frac{-2t^2}{1-t^2}dt$$