$\dfrac {dy}{dx}=\dfrac{4x+5}{3y^2+1}$

$(3y^2+1)~dy = (4x+5)~dx$

$y^3+y = 2x^2 + 5x + C$

we need to find $C$ to satisfy the constraint that the curve passes through $(1,0)$

$0=2(1^2) + 5(1) + C$

$0=2 + 5 + C$

$C=-7$

The implicit form of your curve is thus

$y^3+y=2x^2+5x-7$

This will have an explicit form for $y$ in terms of $x$ but it will be a mess. The implicit form should suffice.