Answer
$-\dfrac{3}{(r-\sqrt 2)^2}+c$
Work Step by Step
Use formula $\int x^{a} dx=\dfrac{x^{(a+1)}}{(a+1)}+c$
where $c$ refers to a constant of proportionality.
Plug $a=r-\sqrt 2$ and $dr =da$
Then, we have $\int \dfrac{6dr}{(r-\sqrt 2)^2}dr=6\int \dfrac{1}{a^3}da=6\int a^{-3} da$
and, $(6)[\dfrac{a^{-3+1}}{-3+1}]+c=-3a^{-2}+c$
Hence, $-3a^{-2}+c=-\dfrac{3}{(r-\sqrt 2)^2}+c$
