$\dfrac {dy}{dx}=\dfrac{4x+5}{3y^2+1}$
$(3y^2+1)~dy = (4x+5)~dx$
$y^3+y = 2x^2 + 5x + C$
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.