Step 1: Simplify the root.
For $x>0$, $\sqrt{x^4+2x^3+2x^2}=\sqrt{x^2(x^2+2x+2)}=x\sqrt{x^2+2x+2}$, so
\[ I=\int\frac{dx}{x\sqrt{x^2+2x+2}}. \]
Step 2: Substitute $x=1/t$.
Then $dx=-\dfrac{1}{t^2}\,dt$, and after substituting and simplifying the inside of the root,
\[ I=-\int\frac{dt}{\sqrt{2t^2+2t+1}}. \]
Step 3: Pull out $\sqrt2$.
\[ I=-\frac{1}{\sqrt2}\int\frac{dt}{\sqrt{t^2+t+\tfrac12}}. \]
Step 4: Complete the square.
$t^2+t+\tfrac12=\left(t+\tfrac12\right)^2+\left(\tfrac12\right)^2$, a form $\sqrt{u^2+a^2}$.
Step 5: Use the standard sinh integral.
Since $\int\dfrac{du}{\sqrt{u^2+a^2}}=\sinh^{-1}\dfrac{u}{a}$,
\[ I=-\frac{1}{\sqrt2}\sinh^{-1}\!\left(\frac{t+\tfrac12}{\tfrac12}\right)+c=-\frac{1}{\sqrt2}\sinh^{-1}(2t+1)+c. \]
Step 6: Return to $x$.
With $t=\dfrac1x$, $2t+1=\dfrac{2}{x}+1=\dfrac{x+2}{x}$, so
\[ I=-\frac{1}{\sqrt2}\sinh^{-1}\!\left(\frac{x+2}{x}\right)+c. \]
\[ \boxed{-\dfrac{1}{\sqrt2}\sinh^{-1}\!\left(\dfrac{x+2}{x}\right)+c} \]