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

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Answer 1 · 2017-07-08T16:22:42

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

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. $(x-2)=t rArr dx=dt.$

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

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

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

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