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
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No need for trigonometric subsitution. Simply let: $\displaystyle u = x - 2 \ \Rightarrow \ du = dx$ So our integral becomes: $\displaystyle 8 \int \frac{1}{(x-2)^2} \ dx = 8 \int \frac{1}{u^2} \ du$
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