solve this 1st order linear equation $$(x+3)^2\frac{dy}{dx}=6-12y-4xy=6-y(12+4x)$$
a. Write it in standard form.
$$\frac{dy}{dx}+\frac{12+4x}{(x+3)^2}y=\frac6{(x+3)^2}$$
b. What is the integrating factor?
$$\frac{12+4x}{(x+3)^2} = \frac14x+\frac34 \implies$$
$$IF=e^{\int{\frac14x+\frac34}}=e^{\frac18x^2+\frac34x}$$
c. Integrate the DE subject to $y(0)=1$.
$$e^{\frac18x^2+\frac34x}\frac{dy}{dx}+ye^{\frac18x^2+\frac34x}(\frac14x+\frac34)=\frac{6e^{\frac18x^2+\frac34x}}{(x+3)^2} \implies$$
$$ye^{\frac18x^2+\frac34x}=\int{\frac{6e^{\frac18x^2+\frac34x}}{(x+3)^2}}+c$$
I have done this problem a few times and now getting kind of stuck. Please check my standard form and integrating factor.
Right now if everything else is correct I cannot solve the integral. Any help appreciated.

$$(x+3)^2\frac{dy}{dx}=6-4y(3+x)$$
$$(x+3)^2\frac{dy}{dx}+4y(3+x)=6$$
Multiply by $(x+3)^2$
$$(x+3)^4y'+4y(3+x)^3=6(x+3)^2$$
$$((x+3)^4y)'=6(x+3)^2$$
Integrate
$$(x+3)^4y=2(x+3)^3+C$$
$$y(x)=\frac 2 {(x+3)}+\frac C {(x+3)^4}$$

As pointed out in the comment this line is not correct
$$\frac{12+4x}{(x+3)^2} = \frac14x+\frac34 $$
This is correct
$$\frac{12+4x}{(x+3)^2} = \frac 4 {x+3}$$
