How to solve x^2 y' + 2xy = arctan(x) - ODE

[Lodve]

I struggle to solve this Differential equation:
$$x^2y' +2xy = arctanx$$
What I did was just divide x^2 on both side of the sign of equality in order to get the same form as a first-order linear diff.equation. After I've done that, I just multiply $$e^(2lnx)$$ on the both side of the equation so that the left side of the equation can be written in the form of $$(u*v)'$$. Now I basically integer on the both side of the equation to remove the derive sign on the left side, but I struggle to integer on the right side of the equation. Can somebody here help me continueing solving this diff.equation? :D

 

[Mark44]

In the original problem, the left side is the derivative of x²y. So you have
d/dx(x²y) = tan^-1(x)
==> x²y = $\int tan^{-1}(x) dx$

You can use integration by parts to integrate the right side.