Answer
$\frac 1 {\sqrt 5}arctan\frac {x-2}{\sqrt 5}+C$
Work Step by Step
$\int \frac 1 {x^2-4x+4+5}dx$
$\int \frac 1 {(x-2)^2+5}dx$
$a=\sqrt 5, u=x-2, du=dx$
$\int \frac {du}{a^2+u^2}$
$\frac 1 a arctan \frac u a +C$
$\frac 1 {\sqrt 5}arctan\frac {x-2}{\sqrt 5}+C$
