Answer
$-\dfrac{3}{(r-\sqrt 2)^2}+c$
Work Step by Step
Since, we know $\int x^{n} dx=\dfrac{x^{n+1}}{n+1}+c$
where $c$ is a constant of proportionality.
Substitute $k=r-\sqrt 2 \implies dr =dk$
Now, $\int \dfrac{6dr}{(r-\sqrt 2)^2}dr=6\int \dfrac{1}{k^3}dk=6\int k^{-3} dk$
or, $=6[\dfrac{k^{-3+1}}{-3+1}]+c$
or, $=-3k^{-2}+c$
plug back in: $k=r-\sqrt 2$
Thus, $=-\dfrac{3}{(r-\sqrt 2)^2}+c$