<s>[INST] Evaluate the integral $\int \:\frac{\sqrt{x^2+2\cdot\:\:x+2}}{x}dx$ $$\int \:\frac{\sqrt{x^2+2\cdot\:\:x+2}}{x}dx$$
I have tried to form a square above i also tried to get the x below under the root
but got nothing
 [/INST] The standard way is to make the substitution
$y=x+\sqrt{x^2+2x+2}$
so that $x=\frac{y^2-2}{2(y+1)}$
$\sqrt{x^2+2x+2}=\frac{y^2+2y+2}{2(y+1)}$
$dx=\frac{y^2+2y+2}{2(y+1)^2}dy$
which changes your integral into an integral of rational function, solvable by partial fraction expansion:
$\int\frac{(y^2+2y+2)^2}{2(y^2-2)(y+1)^2}dy$
It is messy to solve it, but it is definitely doable with the right ammount of determination.
The partial fraction expansion of $\frac{(y^2+2y+2)^2}{2(y^2-2)(y+1)^2}$ is:
$\frac{4}{y^2-2}+\frac{1}{y+1}-\frac{1}{2(y+1)^2}+\frac{1}{2}$
which you integrate term by term.
There might be a easier trick to solve it, but i don't see it.

EDIT: after more thinking i think that one can make the following substitution
$x+1=\tan\theta$ with $\theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$
$\sqrt{x^2+2x+2}=\sqrt{(\tan\theta)^2+1}=\sqrt\frac{1}{\cos^2\theta}=\frac{1}{\cos\theta}$
as $\cos\theta>0$ when $\theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$
Then $dx=\frac{1}{\cos^2\theta} d\theta$
So finally you integrate:
$\int \frac{1}{\cos^2\theta(\sin\theta-\cos\theta)}d\theta$, with $\theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$
From here one might go to the tangent of half angle, but again I see no easy formula.
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