Answer
$-\dfrac{1}{2}\ln(x-2+\sqrt{(x-2)^2+1})+\dfrac{(x-2)\sqrt{(x-2)^2+1}}{2}+4\sqrt{(x-2)^2+1}+4\sinh^{-1}(x-2)+C$
Work Step by Step
Let $I=\int\dfrac{x^2}{\sqrt{(x-2)^2+1}}dx$
Suppose $u=x-2 \implies du=dx$
and $x=u+2 \implies x^2=u^2+4u+4$
Therefore, $I=\int\frac{u^2+4u+4}{\sqrt{u^2+1}}du= \int\dfrac{u^2}{\sqrt{u^2+1}}du+4\int\dfrac{u}{\sqrt{u^2+1}}du+4\int\dfrac{du}{\sqrt{u^2+1}}$
Now, $I=-\dfrac{1}{2}\ln(u+\sqrt{u^2+1})+\dfrac{u\sqrt{u^2+1}}{2}+4\int da+4\sinh^{-1}u+C \\=-\dfrac{1}{2}\ln(u+\sqrt{u^2+1})+\dfrac{u\sqrt{u^2+1}}{2}+4a+4\sinh^{-1}u+C \\ =-\dfrac{1}{2}\ln(x-2+\sqrt{(x-2)^2+1})+\dfrac{(x-2)\sqrt{(x-2)^2+1}}{2}+4\sqrt{(x-2)^2+1}+4 \space \sinh^{-1}(x-2)+C$
