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How do you find the integral of #int 1/(3+(x-2)^2#?

1 Answer

Jul 8, 2017

# 1/sqrt3*arc tan((x-2)/sqrt3)+C.#

Explanation:

Let, #I=int1/(3+(x-2)^2)dx.#

Knowing that, #int1/(t^2+a^2)dt=1/a*arc tan(t/a)+c,# we take

substn. #(x-2)=t rArr dx=dt.#

#:. I=int1/(t^2+sqrt3^2)dt,#

#=1/sqrt3*arc tan (t/sqrt3),#

# rArr I=1/sqrt3*arc tan((x-2)/sqrt3)+C.#

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