What type of functional equation is this? I'm trying to solve the following functional equation
$f\left(x\right)=A\mbox{ exp}\left\{ \int\frac{1}{f\left(x\right)x^{2}+Bx}dx\right\}$
where $f\left(x\right):\mathbb{R}_{+}\rightarrow\mathbb{R}_{\geq0}$, A and B are constants in $\mathbb{R}$.
Does anybody recognize this type of functional equation so I can look up for the solution?
Alternatively, does anybody know how to solve it? Or a suggestion on how to start tackling this problem?
Thank you!
After the initial comments I realized the problem is equivalent to solving the following first-order nonlinear ODE: $f^{\prime}f=\frac{1}{x^{2}}\left(\left(Ax+Bx^{2}\right)f^{\prime}+Bf\right)$ 
This equation seems similar to an Abel differential equation of the second kind, i.e. 
$ff^{\prime}=g\left(x\right)f+h\left(x\right)$ 
although it's not quite the same. If anybody has an idea how to deal with it I would appreciate it!
 A: Differentiate and divide the result by the original expression (on second thought: simply take logarithms, then differentiate) to get a first-order nonlinear ODE :
$$
\ln(f(x))'=1/(f(x)x^2+Bx).
$$
If $B=0$ you get an explicit solution. If $B>0$ there exists more and more slowly yet strictly increasing solutions for all $x\geq 0$
A: First, there is no concept about "indefinite integral equation" , so you should modify the question as $f(x)=Ae^{\int_k^x\frac{1}{f(x)x^2+Bx}dx}$
$\ln\dfrac{f(x)}{A}=\int_k^x\dfrac{1}{f(x)x^2+Bx}dx$
$\dfrac{1}{f(x)}\dfrac{df(x)}{dx}=\dfrac{1}{f(x)x^2+Bx}$ with $f(k)=A$
$f\dfrac{dx}{df}=fx^2+Bx$ with $x(A)=k$
Case $1$: $B=0$
Then $f\dfrac{dx}{df}=fx^2$ with $x(A)=k$
$df=\dfrac{dx}{x^2}$ with $f(k)=A$
$\int_A^fdf=\int_k^x\dfrac{dx}{x^2}$
$[f]_A^f=\left[-\dfrac{1}{x}\right]_k^x$
$f-A=\dfrac{1}{k}-\dfrac{1}{x}$
$f(x)=A+\dfrac{1}{k}-\dfrac{1}{x}$
Case $1$: $B=0$
Then $f\dfrac{dx}{df}=fx^2+Bx$ with $x(A)=k$
$\dfrac{dx}{df}-\dfrac{Bx}{f}=x^2$ with $x(A)=k$
Let $x=\dfrac{1}{y}$ ,
Then $\dfrac{dx}{df}=-\dfrac{1}{y^2}\dfrac{dy}{df}$
$\therefore-\dfrac{1}{y^2}\dfrac{dy}{df}-\dfrac{B}{fy}=\dfrac{1}{y^2}$ with $y(A)=\dfrac{1}{k}$
$\dfrac{dy}{df}+\dfrac{By}{f}=-1$ with $y(A)=\dfrac{1}{k}$
I.F. $=e^{\int\frac{B}{f}df}=e^{B\ln f}=f^B$
$\therefore\dfrac{d(f^By)}{df}=-f^B$
$f^By=-\int f^B~df$
$f^By=\begin{cases}-\dfrac{f^{B+1}}{B+1}+C&\text{when}~B\neq0,-1\\-\ln f+C&\text{when}~B=-1\end{cases}$
$y(A)=\dfrac{1}{k}$ :
$\dfrac{A^B}{k}=\begin{cases}-\dfrac{A^{B+1}}{B+1}+C&\text{when}~B\neq0,-1\\-\ln A+C&\text{when}~B=-1\end{cases}$
$C=\begin{cases}\dfrac{A^B}{k}+\dfrac{A^{B+1}}{B+1}&\text{when}~B\neq0,-1\\\dfrac{1}{kA}+\ln A&\text{when}~B=-1\end{cases}$
$\therefore f^By=\begin{cases}\dfrac{A^B}{k}+\dfrac{A^{B+1}}{B+1}-\dfrac{f^{B+1}}{B+1}&\text{when}~B\neq0,-1\\\dfrac{1}{kA}+\ln A-\ln f&\text{when}~B=-1\end{cases}$
$\dfrac{f^B}{x}=\begin{cases}\dfrac{A^B}{k}+\dfrac{A^{B+1}}{B+1}-\dfrac{f^{B+1}}{B+1}&\text{when}~B\neq0,-1\\\dfrac{1}{kA}+\ln A-\ln f&\text{when}~B=-1\end{cases}$
$x=\begin{cases}\dfrac{1}{\dfrac{A^B}{kf^B}+\dfrac{A^{B+1}}{(B+1)f^B}-\dfrac{f}{B+1}}&\text{when}~B\neq0,-1\\\dfrac{1}{\dfrac{f}{kA}+f\ln A-f\ln f}&\text{when}~B=-1\end{cases}$
