Inconsistent solution to differential equation $Q.$
Let f(x) be a continuous function satisfying the following differential equation-
$$f(x)=(1+x^2)(1+\int_0^x\frac{f^2(t)dt}{1+t^2})$$
$$\text{Find  f(1)}$$
My Work-
1)Putting $x=0$ in the equation we get $f(0)=1$
2)Dividing by $1+x^2$ and differentiating w.r.t. $x$ we get-
$$(\frac{y}{1+x^2})'=\frac{y^2}{1+x^2}\qquad\text{∴ y=f(x)}$$
Simplifying-
$$y'(1+x^2)-2xy=y^2(1+x^2)$$
so either $f(x)=0$ or
$$\frac{-dy(1+x^2)}{y^2}+\frac{2xdx}{y}=(-1-x^2)dx$$
$$\frac{d}{dx}(\frac{1+x^2}{y})=\frac{d}{dx}(-x-\frac{x^3}{3})$$
$$\frac{1+x^2}{y}=-x-\frac{x^3}{3}+c$$
Using $y(0)=1$ we get $c=1$ and hence $$f(1)=-6$$
My issue- looking at the question, no matter the value of f(x), the R.H.S. of the equation must always be positive (squared) however the answer is coming to be negative. Am I missing something? Or is there any error in the question? Or is $f(x)=0$ the only acceptable solution?

$$\frac{f(x)}{1+x^2}=1+\int_{0}^{x} \frac{f^2(t)}{1+t^2}dt \implies f(0)=1$$
D. w.r.t. $x$, using Lebnitz, to get
$$f'(x)-\frac{2x}{1+x^2}f(x)=f^2(x)\implies f^{-2}f'-\frac{2x}{1+x^2}\frac{1}{f}=1$$
This is Bernoulli equation. Let $1/f=v$, then we get
$$\frac{dv}{dx}+\frac{2x}{1+x^2}v=-1$$
This is linear equation, whose integrating factor is $I=\exp[\int \frac{2x}{1+x^2}dx]\implies I= (1+x^2)$
$$v=(1+x^2)^{-1} \int -1(1+x^2) dx + C (1+x^2)^{-1}.$$
$$\implies v=\frac{-x-x^3/3}{1+x^2}+C(1+x^2)^{-1}=\frac{1}{f}$$
Used $f(0)=1 \implies C=1.$ Finally, $$f(x)=\frac{1+x^2}{1-x-x^3/3}.$$
So $f(1)=-6.$ The plot of $f(x)$ is given below wherein $f(x)>0$only in (0,0.8177.. ), then there is a singularity at $x=0.8177..$ and it becomes negative. The answer to OP's interesting question could lie in the non-linearity of this ODE.

