Answer
See below
Work Step by Step
Given: $y''+6y'+9y=\frac{2e^{-3x}}{x^2+1}$
In this case, the auxiliary polynomial of the associated homogeneous equation is
$P(r) = r^2 +6r+9 =0\\
\rightarrow (r+3)^2=0\\
\rightarrow r=-3$
so that three linearly independent solutions are:
$y_1(x)=e^{-3x}\\
y_2(x)=xe^{-3x}$
A particular solution is:
$y_p(x)=u_1(x)e^{-3x}+u_2(x)xe^{-3x}$
Obtain $u_1'(x)e^{-3x}+u_2'(x)xe^{-3x}=0\\
u_1'(x)e^{-3x}+u_2'(x)(xe^{-3x})'= \frac{2e^{-3x}}{x^2+1}$
Consequently $u_1'=-\frac{2x}{x^2+1}\\
u_2'=\frac{2}{x^2+1}$
By integrating, we obtain
$u_1=\int -\frac{2x}{x^2+1}dx=-\ln (x^2+1)+C_2\\
u_2=\int \frac{2}{x^2+1}dx=2\tan^{-1}x+C_1 $
where $C_1,C_2$ are integration constants.
The general solution to the given differential equation is therefore:
$y(x)=(2\tan^{-1}x+C_1)e^{-2x}+[-\ln (x^2+1)+C_2]xe^{-3x}$
