Need help solving 2 different differential equations (partially solved, but stuck)

I've been working on these for a WHILE, and they end up giving me integrals which seem pretty much impossible to solve or crazy unrealistic answers. If I'm doing the first few steps right, then can someone walk me through how to solve the integrals? Or, is there an easier way to solve the differential equations?:

1) solve (y-2x)dx + (3x-y)dy = 0

2) solve 3xydx + (x^2 + y^2)dy = 0

for number (1), I substituted y = ux and dy = udx + xdu

after some simplifying, I got:

(-u^2 + 4u - 2)xdx = -x^2(3-u)du

-dx/x = (u-3)/(u^2 -4u +2)du

-lnx = int[(u-3)du/(u^2 -4u +2)]

after trying all kinds of crazy weird stuff, including trig substitution, I ended up getting:

-lnx = (1/2)ln[(y/x)^2 -4y/x +2) - (1/4)arctan[{(y/x)-2}/sqrt(2)] - (1/8)sin(2arctan[{(y/x) - 2)}\sqrt(2)]

this just seems like a ridiculous answer, and I'm pretty sure I went wrong somewhere along the line. If someone could check through this ASAP and let me know where I went wrong, I'd greatly appreciate it!

2) Its a similar problem to #1. I subsituted y=ux and dy=udx + xdu

The integral I ended up getting was int[-dx/x] = int[(1+u^2)/(u^3 + 4u)]

The right side is what I'm having trouble with.

I tried seperating it into: (u)/(u^2 + 4) and 1/(u^3 + 4u)

The first part I solved through "u substitution" getting (1/2)ln(4 + u^2)

the 2nd part (with the 1 on top) I tried solving through partial fractions, but when I used partial fractions, I got (1/4) for A and 0 for both B and C, making it equal to 1/(4u) which I know is wrong. (Headbang)

Any help is greatly appreciated, thank you!