Verified answer

If you multiply through by

x³ dt

This recasts your equation as

(x^5+4tx^4) dt +(5tx^4+8t²x^3) dx=0

which is just

d(tx^5) + d(2t²x^4) = 0

and directly integrable

PS. I like to multiply through with that dt right at the beginning. If you're not used to it, you can save it for the end. I like clearing the denominator earlier and to my mind it is more symmetric. If you keep the dt on the bottom you have:

d(tx^5)/dt + d(2t²x^4)/dt = 0

And now if you want to integrate, you really should clear out those dt's.

x² + 4tx + (5tx + 8t²)((dx)/(dt))=0

dx/dt = -(x² + 4tx)/(5tx + 8t²)

Let x = vt. Then dx/dt = v + t(dv/dt)

Thus (1) becomes

v + t(dv/dt) = -((vt)² + 4t(vt))/(5t(vt) + 8t²)

v + t(dv/dt) = - (v^2 + 4v)/(5v + 8)

t(dv/dt) = - (v^2 + 4v)/(5v + 8) - v = - (6v^2 + 12v)/(5v + 8)

[(5v + 8)/(6v^2 + 12v)] dv + [1/t] dt = 0

(1/6)∫[(5v + 8)/(v^2 + 2v)] dv + ∫[1/t] dt = 0

(1/6)∫[{(2v + 2)+(3v + 6)}/(v^2 + 2v)] dv +ln(t) = ln(c)

(1/6) ∫[(2v + 2)/(v^2 + 2v)] dv + (1/6) ∫[3(v + 2)/{v(v + 2)}] dv + ln(t) = ln(c)

(1/6) ∫[d(v^2 + 2v)/(v^2 + 2v)] + (1/2) ∫[dv/v] + ln(t) = ln(c)

(1/6) ln(v^2 + 2v) + (1/2) ln(v) + ln(t) = ln(c)

ln[(v^2 + 2v)^(1/6)] + ln[v^(1/2)] + ln(t) = ln(c)

ln[(v^2 + 2v)^(1/6)*v^(1/2)*t] = ln(c)

(v^2 + 2v)^(1/6)*v^(1/2)*t = c

Substituting for v

{(x/t)^2 + 2(x/t)}^(1/6)*(x/t)^(1/2)*t = c

(tx^2 + 2xt^2)^(1/6)*x^(1/2) = c