Answer
$\displaystyle \arctan x-\frac{1}{2(x^{2}+1)}+C$
Work Step by Step
$ I=\displaystyle \int\frac{x^{2}+x+1}{(x^{2}+1)^{2}}dx$
$ \displaystyle \frac{x^{2}+x+1}{(x^{2}+1)^{2}}=\frac{Ax+B}{x^{2}+1}+\frac{Cx+D}{(x^{2}+1)^{2}}$
$x^{2}+x+1=(Ax+B)(x^{2}+1)+Cx+D$
$x^{2}+x+1=Ax^{3}+Ax+Bx^{2}+B+Cx+D$
$x^{2}+x+1=Ax^{3}+Bx^{2}+x(A+C)+(B+D)$
$A=0, B=1$
$A+C=1\Rightarrow C=1$
$B+D=1\Rightarrow D=0$
$ \displaystyle \frac{x^{2}+x+1}{(x^{2}+1)^{2}}=\frac{1}{x^{2}+1}+\frac{x}{(x^{2}+1)^{2}}$
$\displaystyle \int\frac{x}{(x^{2}+1)^{2}}dx=\left[\begin{array}{l}
t=x^{2}+1\\
dt=2xdx
\end{array}\right]=\frac{1}{2}\int t^{-2}dt$
$=-\displaystyle \frac{1}{2t}=-\frac{1}{2(x^{2}+1)}$
$I=\displaystyle \arctan x-\frac{1}{2(x^{2}+1)}+C$
