Answer
$$\int\frac{y^2+2y+1}{(y^2+1)^2}dy=\tan^{-1}y-\frac{1}{y^2+1}+C$$
Work Step by Step
$$A=\int\frac{y^2+2y+1}{(y^2+1)^2}dy$$
1) Express the integrand as a sum of partial fractions:
Here we have both an irreducible quadratic factor and a repeated linear factor,
$$\frac{y^2+2y+1}{(y^2+1)^2}=\frac{Ay+B}{y^2+1}+\frac{Cy+D}{(y^2+1)^2}$$
Clear fractions:
$$(Ay+B)(y^2+1)+(Cy+D)=y^2+2y+1$$ $$Ay^3+Ay+By^2+B+Cy+D=y^2+2y+1$$ $$Ay^3+By^2+(A+C)y+B+D=y^2+2y+1$$
Equating coefficients of corresponding powers of $y$, we get
$A=0$ and $B=1$
$A+C=2$, so $C=2$
$B+D=1$, so $D=0$
Therefore,
$$\frac{y^2+2y+1}{(y^2+1)^2}=\frac{1}{y^2+1}+\frac{2y}{(y^2+1)^2}$$
2) Evaluate the integral:
$$A=\int\frac{dy}{y^2+1}+\int\frac{2y}{(y^2+1)^2}dy$$ $$A=\tan^{-1}y+\int\frac{1}{(y^2+1)^2}d(y^2+1)$$ $$A=\tan^{-1}y-\frac{1}{y^2+1}+C$$
