1. ## integration with arctan

integrate:

8 / (x-2)^2 dx

i tried to expand to

8 / (x^2 + 4 -4x) and then substitute x = tan theta .... but i got nowhere

2. No need for trigonometric subsitution. Simply let: $u = x - 2 \ \Rightarrow \ du = dx$

So our integral becomes: $8 \int \frac{1}{(x-2)^2} \ dx = 8 \int \frac{1}{u^2} \ du$