Originally Posted by
teuthid This is in essence exactly the same procedure that the others have used, but with slightly different symbols (my old calc teacher used to call it "tricky" substitution)...
choose u=x-4. This means that du=dx, but it also means that u+4=x. As a result we have the following integral:
$\displaystyle \int \frac{2(u+4)}{u} du$
$\displaystyle
=\int \frac{2u}{u} +\frac{8}{u} du$
$\displaystyle
=2\int du + 8\int \frac{1}{u} du$
$\displaystyle =2u+8\ln{u}+c$
$\displaystyle =2(x+4)+8\ln{(x-4)}+c$
$\displaystyle =2x+8\ln{(x-4)}+C$