Integrate 1/ ( x^2 + y^2) ^3/2 with respect to dy

Can I use trig sub.

Yes, use a trig. substitution.

Let y = x tan t, dy = x sec^2(t) dt.

So, ∫ dy/(x^2 + y^2)^(3/2)
= ∫ (x sec^2(t) dt) / [x^2 + x^2 tan^2(t)]^(3/2)
= ∫ (x sec^2(t) dt) / [x^2 sec^2(t)]^(3/2)
= ∫ (x sec^2(t) dt) / (x^3 sec^3(t))
= ∫ (1/x^2) cos t dt
= (1/x^2) sin t + C
= (1/x^2) * y/sqrt(x^2 + y^2) + C, since tan t = y/x and 'sohcahtoa'.

I hope this helps!