Question

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

Answer 1

`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.`