Definite integral $\int_{0}^{1}( \frac{1}{\sqrt{x\ln x}}+\sqrt{\frac{\ln x}{x}}) dx$ $$\int_{0}^{1}\left( \frac{1}{\sqrt{x\ln x}}+\sqrt{\frac{\ln x}{x}}\right) dx$$
Here is what I tried and failed
$$\int_{0}^{1}\frac{1}{\sqrt{x} }\left(\frac{1}{\sqrt{\ln x}}+\sqrt{\ln x}\right)dx$$
$$\int_{0}^{1} \frac{1}{\sqrt{x}}\left( \frac{1 + \ln x}{\sqrt{\ln x}}\right)dx$$
After that, I got stuck.
 A: $$I={\displaystyle\int}\dfrac{\sqrt{\ln\left(x\right)}}{\sqrt{x}}\,\mathrm{d}x+{\displaystyle\int}\dfrac{1}{\sqrt{x}\sqrt{\ln\left(x\right)}}\,\mathrm{d}x$$
For the integral${\displaystyle\int}\dfrac{\sqrt{\ln\left(x\right)}}{\sqrt{x}}\,\mathrm{d}x$ Integrate by parts 
$$I=2\sqrt{x}\sqrt{\ln\left(x\right)}-{\displaystyle\int}\class{steps-node}{\cssId{steps-node-1}{\dfrac{1}{\sqrt{x}\sqrt{\ln\left(x\right)}}}}\,\mathrm{d}x+{\displaystyle\int}\dfrac{1}{\sqrt{x}\sqrt{\ln\left(x\right)}}\,\mathrm{d}x$$
The integral 
${\displaystyle\int}\dfrac{1}{\sqrt{x}\sqrt{\ln\left(x\right)}}\,\mathrm{d}x$
cancels, leaving only:
$$\fbox { $I=2\sqrt{x}\sqrt{\ln\left(x\right)}+C $}$$
$x$ goes from $0$ to $1$
$$\lim_{x \to 0^+}\sqrt x \sqrt \ln x=0 $$
Thus $$I=2\sqrt 1 \cdot 0-0=0$$ 
A: The first thing to do here is to realize that we are taking square roots of negative numbers, so we need to proceed carefully, so as not to be led astray by ostensible identities like $1/\sqrt z=\sqrt{1/z}$, which is correct for positive reals but problematic for negative reals. The best thing to do (in my experience) is to reexpress things so as to get rid of the square root symbol altogether.
One way to do this is to let $x=e^{-u^2}$ with $0\le u\lt\infty$. We see that
$${1\over\sqrt{x\ln x}}+\sqrt{\ln x\over x}={1\over\sqrt{-u^2e^{-u^2}}}+\sqrt{-u^2\over e^{-u^2}}={u\sqrt{-1}\over e^{-u^2/2}}+{1\over ue^{-u^2}\sqrt{-1}}=ie^{u^2/2}\left(u-{1\over u} \right)$$
(Note the crucial minus sign in the final step!)
With $x=e^{-u^2}\implies dx=-2ue^{-u^2}du$, we find
$$\int_0^1\left({1\over\sqrt{x\ln x}}+\sqrt{\ln x\over x}\right)dx=-2i\int_\infty^0e^{u^2/2}\left(u-{1\over u} \right)ue^{-u^2}du=2i\int_0^\infty(u^2-1)e^{-u^2/2}du$$
Finally, integration by parts (with $u=u$ and $dv=ue^{-u^2/2}du$, so that $v=-e^{-u^2/2}$) tells us
$$\int_0^\infty u^2e^{-u^2/2}du=-ue^{-u^2/2}\big|_0^\infty+\int_0^\infty e^{-u^2/2}du=(0-0)+\int_0^\infty e^{-u^2/2}du=\int_0^\infty e^{-u^2/2}du$$
so that
$$\int_0^\infty(u^2-1)e^{-u^2/2}du=\int_0^\infty u^2e^{-u^2/2}du-\int_0^\infty e^{-u^2/2}du=\int_0^\infty e^{-u^2/2}du-\int_0^\infty e^{-u^2/2}du=0$$
Remark: I've spelled things out in more steps than may be necessary. The key steps are the appearance of the negative sign in the first display and the separation of $u^2$ from $-1$ for integration by parts to get to the cancelation of integrals. A close reading of Deepesh Meena's answer should reveal that the same two steps are in effect taken there as well, without the change of variable, but I have to admit I haven't read that answer closely enough to see how it avoids making a $1/\sqrt{z}=\sqrt{1/z}$ type error when it manipulates square roots of the things that are negative.
