If the solution $$y = y(x)$$ of the differential equation $$(x^4 + 2x^3 + 3x^2 + 2x + 2)dy - (2x^2 + 2x + 3)dx = 0$$ satisfies $$y(-1) = -\frac{\pi}{4}$$, then y(0) is equal to:

Write the differential equation in the separable form:

$$\bigl(x^{4}+2x^{3}+3x^{2}+2x+2\bigr)\,dy-\bigl(2x^{2}+2x+3\bigr)\,dx=0$$
$$\Longrightarrow\;dy=\frac{2x^{2}+2x+3}{x^{4}+2x^{3}+3x^{2}+2x+2}\,dx$$

Factor the quartic in the denominator:

$$x^{4}+2x^{3}+3x^{2}+2x+2 =(x^{2}+1)(x^{2}+2x+2)$$

Hence

$$\frac{2x^{2}+2x+3}{x^{4}+2x^{3}+3x^{2}+2x+2} =\frac{2x^{2}+2x+3}{(x^{2}+1)(x^{2}+2x+2)}$$

Resolve the integrand into partial fractions.
Assume

$$\frac{2x^{2}+2x+3}{(x^{2}+1)(x^{2}+2x+2)} =\frac{Ax+B}{x^{2}+1}+\frac{Cx+D}{x^{2}+2x+2}$$

Clearing denominators and equating coefficients gives the system

$$\begin{aligned} A+C &=0,\\ 2A+B+D &=2,\\ A+2B &=2,\\ 2B+D &=3. \end{aligned}$$

Solving, we obtain $$A=0,\;B=1,\;C=0,\;D=1.$$
Therefore

$$\frac{2x^{2}+2x+3}{(x^{2}+1)(x^{2}+2x+2)} =\frac{1}{x^{2}+1}+\frac{1}{x^{2}+2x+2}.$$

Integrate term by term:

$$\int\frac{1}{x^{2}+1}\,dx=\arctan x,$$
Let $$u=x+1,$$ so $$\int\frac{1}{x^{2}+2x+2}\,dx =\int\frac{1}{u^{2}+1}\,du=\arctan u=\arctan(x+1).$$

Thus the general solution is

$$y(x)=\arctan x+\arctan(x+1)+C.$$

Use the initial condition $$y(-1)=-\frac{\pi}{4}:$$

$$-\frac{\pi}{4}=y(-1)=\arctan(-1)+\arctan 0+C =\Bigl(-\frac{\pi}{4}\Bigr)+0+C,$$
so $$C=0.$$

Hence $$y(x)=\arctan x+\arctan(x+1).$$

Finally, evaluate at $$x=0:$$

$$y(0)=\arctan 0+\arctan 1 =0+\frac{\pi}{4} =\frac{\pi}{4}.$$

Therefore, $$y(0)=\frac{\pi}{4},$$ which corresponds to Option D.