$\int_{0}^{x^3+1}\frac{f'(t)}{1+f(t)}dt=\ln(x)$ Solve for $f(x)$.
$$\int_{0}^{x^3+1}\frac{f'(t)}{1+f(t)}dt=\ln(x)$$
I attempted a simpler version without the $x^3+1$ bound to just get a sense for the problem.
$$\int_0^x \frac{f'(t)}{1+f(t)} dt = \ln(x)$$
Using the Fundamental Theorem of Calculus Part I, we can take the derivative on both sides of the equation to clean up the integral symbols and eliminate the $t$ variables.
$$\Rightarrow\frac{d}{dx}\left[\int_0^x \frac{f'(t)}{1+f(t)} dt = \ln(x)\right]\\\Rightarrow \frac{f'(x)}{1+f(x)}=\frac{1}{x}$$
Assume $y = f(x)$.$$\frac{dy}{dx}=\frac{1+y}{x}$$
The following differential equation can be solved by seperation and integrating both sides of the equation, which leads us to $y=f(x)=x-1$ $$\Rightarrow\int\left[\frac{dy}{y+1}=\frac{dx}{x}\right]\\\Rightarrow\ln|y+1|=\ln|x|\\\Rightarrow y=f(x)=x-1$$
Now going back to our original problem, we again take the derivative on both sides. This time we have to implement the chain rule for FTC Part I to work and then solve our differential equation (which proved to be difficult for me).
$$\Rightarrow\frac{d}{dx}\left[\int_{0}^{x^3+1}\frac{f'(t)}{1+f(t)}dt=\ln(x)\right]\\\Rightarrow\frac{3x^2f'(x^3+1)}{1+f(x^3+1)}=\frac{1}{x}\\\Rightarrow 3x^3f'(x^3+1)=1+f(x^3+1)$$
Now I wasn't really sure what to do from this point onwards. I could just assume $x^3+1$ as an independent variable and then just solve the differential equation, but I'm just not sure if that's the correct approach.      I would appreciate if you not give me the entire answer but rather a prompt in the right direction. Thanks in advance (:

$$\\\Rightarrow 3x^3f'(x^3+1)=1+f(x^3+1)\implies x(3x^2)f'(x^3+1)=1+f(x^3+1)$$
$$\implies \frac{3x^2f'(x^3+1)}{1+f(x^3+1)}=\frac{1}{x}$$
Now integrate both sides with respect $x$ and also note that that the derivative of denominator on LHS is written in numerator, hence using substitution and some manipulation , we get
$$f(x^3+1)=Ax-1$$
where $A$ is a constant and therefore
$$\boxed {f(x)=A(x-1)^{\frac{1}{3}}-1}$$
