<s>[INST] What is this integral $\int {\sqrt{\frac{x+1}{x}}}\:dx$? I tried a lot of different approaches but found this problem very hard. 
So can you help me with this integral?
$$\int {\sqrt{\frac{x+1}{x}}}\:dx$$
Thanks.
 [/INST] I have finally understood what Surb meant in his answer and I'm posting it here for the benefit of future readers. I can't post it as a comment due to space restrictions. 
$\int{\sqrt{\frac{x+1}{x}}}$
Let u = $\sqrt{\frac{x+1}{x}}$ and v = $\frac{x+1}{x} = 1 + \frac{1}{x}$
$\therefore u = \sqrt{v}$
$\frac{dv}{dx} = \frac{-1}{x^{2}}$
$\frac{du}{dx} = \frac{du}{dv} \frac{dv}{dx} = \frac{1}{2 \sqrt{v}} \frac{-1}{x^{2}} = \frac{-1}{2ux^{2}}$
$u^{2} = 1 + \frac{1}{x} \therefore u^{2} - 1 = \frac{1}{x} \therefore x^{2} = \frac{1}{( u^{2} - 1)^{2}}$
$\therefore \frac{du}{dx} = \frac{- ( u^{2} - 1)^{2}}{2u}$
$\therefore dx = \frac{-2u}{ ( u^{2} - 1)^{2}} du$
$\therefore \int{\sqrt{\frac{x+1}{x}}}dx = \int{ \frac{-2u^{2}}{ ( u^{2} - 1)^{2}}}du = -2 \int{ \frac{u^{2}}{ ( u^{2} - 1)^{2}}}du = -2 \int{ u \times \frac{u}{ ( u^{2} - 1)^{2}}}du$
Using rule $\int{a b'} = ab - \int{b a'}$ where a = u and b' = $\frac{u}{ ( u^{2} - 1)^{2}}$
Let v = $u^{2} - 1$ then
b = $\int{b'} = \int{ \frac{u}{ ( u^{2} - 1)^{2}}}du = \frac{1}{2} \int{\frac{1}{v^{2}}}dv = \frac{-1}{2(u^{2} - 1)}$
$\int{a b'} = u \times \frac{-1}{2(u^{2} - 1)} - \int{ \frac{-1}{2(u^{2} - 1)}}du = \frac{-u}{2(u^{2} - 1)} + \frac{1}{2} \int{\frac{1}{u^{2} - 1}}du$
$\frac{1}{u^{2} - 1} = \frac{1}{u+1} \frac{1}{u-1} = \frac{1}{2(u-1)} - \frac{1}{2(u+1)}$
$\therefore \int{\frac{1}{u^{2} - 1}}du = \int{\frac{1}{2(u-1)} - \frac{1}{2(u+1)}}du = \frac{log(u-1) - log(u+1)}{2}$
$\therefore \int{a b'} = \frac{-u}{2(u^{2} - 1)} + \frac{log(u-1) - log(u+1)}{4}$
$\therefore \int{\sqrt{\frac{x+1}{x}}}dx = \frac{u}{u^{2} - 1} + \frac{log(u+1) - log(u-1)}{2} = x \sqrt{\frac{x+1}{x}} + \frac{log( \sqrt{\frac{x+1}{x}}+1 ) - log( \sqrt{\frac{x+1}{x}} - 1)}{2}$
Drumroll please. And thanks Surb for the hint. You are awesome. You really helped Jupiter. 
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