Answer
$\tan^{-1}x-\frac{1}{2(x^2+1)}+C$
Work Step by Step
$\int\frac{x^2+x+1}{(x^2+1)^2}dx$ = $\int\left(\frac{Ax+B}{x^2+1}+\frac{Cx+D}{(x^2+1)^2}\right)dx$
$x^2+x+1$ = $(Ax+B)(x^2+1)+(Cx+D)$
$x^2+x+1$ = $(Ax^3+Bx^2+Ax+B)+(Cx+D)$
$0$ = $A$
$1$ = $B$
$1$ = $A+C$
$1$ = $B+D$
$A$ = $0$, $B$ = $1$, $C$ = $1$, $D$ = $0$
So
$\int\frac{x^2+x+1}{(x^2+1)^2}dx$ = $\int\left[\frac{1}{x^2+1}+\frac{x}{(x^2+1)^2}\right]dx$ = $\tan^{-1}x-\frac{1}{2(x^2+1)}+C$
