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

Rationalise the numerator.
$\displaystyle \int \frac{\sqrt{x^2+2x+2}}{x} \, \text{d}x = \int \frac{x^2+2x+2}{x\sqrt{x^2+2x+2}}\, \text{d}x$
Split apart as follows:
$\displaystyle \int \frac{x^2+2x+2}{x\sqrt{x^2+2x+2}}\, \text{d}x = \int \frac{x+1}{\sqrt{x^2+2x+2}} \, \text{d}x + \int \frac{\text{d}x}{\sqrt{x^2+2x+2}} + \int \frac{2\,\text{d}x}{x\sqrt{x^2+2x+2}}$
The first is an application of the reverse chain rule, the second is an application of the inverse hyperbolic sine formula, and the third requires the substitution $\displaystyle u = \frac{2}{x}$.
$\displaystyle \int \frac{\sqrt{x^2+2x+2}}{x} \, \text{d}x = \sqrt{x^2+2x+2} + \log{\left(x+1+\sqrt{x^2+2x+2} \right)} - \int \frac{\frac{4 \text{d}u}{u^2}}{\frac{2}{u} \sqrt{\frac{4}{u^2}+\frac{4}{u}+2}} $
$\displaystyle \int \frac{\frac{4 \text{d}u}{u^2}}{\frac{2}{u} \sqrt{\frac{4}{u^2}+\frac{4}{u}+2}} = \int \frac{\sqrt{2}\,\text{d}u}{\sqrt{u^2+2u+2}} = \sqrt{2} \log{\left( u+1+\sqrt{u^2+2u+2} \right)}$
$\displaystyle \int \frac{\sqrt{x^2+2x+2}}{x} \, \text{d}x = \sqrt{x^2+2x+2} + \log{\left(x+1+\sqrt{x^2+2x+2} \right)} - \sqrt{2} \log{\left( \frac{2}{x}+1+\sqrt{\frac{4}{x^2}+\frac{4}{x}+2} \right)} $
$\displaystyle = \sqrt{x^2+2x+2} + \log{\left(x+1+\sqrt{x^2+2x+2} \right)} +\sqrt{2}\log{x} - \sqrt{2}\log{\left( x+2+\sqrt{2}\sqrt{x^2+2x+2} \right)}$
