How to treat dy/dx when integrating

Hello all!

I'm feeling pretty sheepish. I finished calc 3 two years ago and am just starting diff EQ. The first section is kicking my butt pretty hard.

Here's my current homework problem:

y'=(x-3)(y^2+1), y(0)=1

Here's what I've done so far:

y'/(y^2+1)=(x-3)

Integral of (y'/(y^2+1)) = integral of (x-3)

arctan(y)=(x^2/2-3x+C)

My issue is in how I treat Y prime when integrating. I know the answer I have is correct so far - Wolfram Alpha says so - but I've lost why y' (AKA dy/dx) is being treated as "1". I understand that it is, I've lost why it is.

My book is concise to a fault, so I need some help understanding this because it's a basic building block that I should remember but don't. I forgot that integration as well - I had to look up what it took to get an arctan after integrating!

(Answer is: arctan(1) = 0-0+C, C= pi/4

arctan(y) = x^2/2 -3x + pi/4

Book simplifies it to: arctan(y) = (2x^2-12x+pi)/4 )

Re: How to treat dy/dx when integrating

You should turn y' into dy/dx to separate variables.

From here you cannot integrate the right hand side since there is no dx in the integral.

Instead you should treat the problem like this

Now you can integrate both sides