Question

integrate 1/(x-2)(x+4) when x>2

Answer 1

seoarate fractions:
\[\frac{A}{x-2}+\frac{B}{x+4}=\frac{A(x+4)+B(x-2)}{(x-2)(x+4)}=\frac{Ax+4A+Bx-2B}{(x-2)(x+4)}\]
so A+B=0 and 4A-2B=1
solving this system you faind that A=1/6 and B=-1/6.
So your integral becomes:
\[\frac{1}{6}\int\limits\frac{dx}{(x-2)}-\frac{1}{6}\int\limits\frac{dx}{x+4}=1/6Ln(x-2)-1/6Ln(x+4) +C\]

Answer 2

It also needs to be evaluated on the interval (2,infinity)

Answer 3

it's same, as long the (x-2)>0