Definite integral of $\frac{1-x}{x}$ vs $\sqrt{\frac{1-x}{x}}$ I found myself intrigued by the result of the definite integrals of $\frac{1-x}{x}$ and $\sqrt{\frac{1-x}{x}}$ respectively. Both functions are seemingly similar, with vertical asymptotes at $x=0$. However the area bounded between the two and the $x$-axis for $x=[0,1]$ is clearly different. Let $f(x)=\frac{1-x}{x}$:
                    

and so we have:
$$
\begin{align}
&\int\limits^{1}_{0}f(x)\,dx=\infty\\
&\int\limits^{1}_{0}f(x)^{1/2}\,dx=\frac{\pi}{2}
\end{align}
$$
Now, since $\displaystyle\lim\limits_{x\to0}\,f(x)=\lim\limits_{x\to0}\,f(x)^{1/2}=\infty$ shouldn't  the two both evaluate to infinity? Even though it is clear that $\int\limits^{1}_{0}f(x)^{1/2}\,dx$ will be less than $\int\limits^{1}_{0}f(x)\,dx$, how can that be that one evaluates to infinity while the other to merely $\approx1.57\!\!\ldots$?
Has this something to do with this? Or perhaps with the fact that $\ln{x}$, which forms a part of the indefinite integral for $f(x)$, isn't defined for $0$?
 A: Well, this is not that different from what happens with some infinite series: $$\sum_{n=1}^\infty\frac{1}{n}=\infty$$ $$\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}$$ and, in fact, the explanation is somehow similar.
A: One has that
$$\int_0^1 \frac {1-x} {x } dx =\left. \log x -x \right|_0^1\to \infty$$
Whereas
$$\int_0^1 \sqrt{\frac {1-x} {x }} dx =$$
$$x=\sin^2 \theta$$
$$dx=2 \sin \theta \cos \theta$$
$$\int_0^{\frac{\pi }{2}} {\sqrt {\frac{{1 - {{\sin }^2}\theta }}{{{{\sin }^2}\theta }}} } 2\sin \theta \cos \theta  = 2\int_0^{\frac{\pi }{2}} {{{\cos }^2}\theta }  = 2\frac{\pi }{4}=\frac{\pi }{2}$$
There is not much more explanation than that. One evaluates in terms of the logarithm, so it diverges. For the other, an idea would be to look at inverses.
The inverse of the latter is $$\frac{1}{{1 + {x^2}}} = y$$ which is well beaved in the real realm. Actually, you're computing 
$$\int\limits_0^\infty  {\frac{1}{{1 + {x^2}}}dx}  = \int_0^1 {\sqrt {\frac{{1 - x}}{x}} } dx = \frac{\pi }{2}$$
