$\int_{0}^{x^3+1}\frac{f'(t)}{1+f(t)}dt=\ln(x)$ 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 (:
 A: Put $s=f(t)$ in the integral. You get $\int_{f(0)} ^{f(x^{3}+1)} \frac 1 {1+s} ds=\ln x$. So $\ln [1+f(x^{3}+1)]-\ln [1+f(0)]=\ln x$. Can you continue from here?
A: $$\\\Rightarrow 3x^3f'(x^3+1)=1+f(x^3+1)\implies x(3x^2)f'(x^3+1)=1+f(x^3+1)$$
$$\implies \frac{3x^2f'(x^3+1)}{1+f(x^3+1)}=\frac{1}{x}$$
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
$$\boxed {f(x)=A(x-1)^{\frac{1}{3}}-1}$$
A: Notice that $$\frac{f'(t)}{f(t)+1}=[\ln\circ(1+f)]'(t),$$ so the equation is equivalent to $$\int_0^{x^3+1}[\ln\circ(1+f)]'(t)\,\mathrm{d}t=\ln(x).$$ By the fundamental theorem of calculus, $$\int_0^{x^3+1}[\ln\circ(1+f)]'(t)\,\mathrm{d}t=\ln[1+f(x^3+1)]-\ln[f(0)],$$ so $$\ln[1+f(x^3+1)]-\ln[1+f(0)]=\ln\left[\frac{1+f(x^3+1)}{1+f(0)}\right]=\ln(x).$$ This is equivalent to $$\frac{1+f(x^3+1)}{1+f(0)}=x.$$ This is equivalent to $$f(x^3+1)=[1+f(0)]\cdot{x}-1.$$ Let $y=x^3+1,$ hence $x=\sqrt[3]{y-1}.$ Therefore, $$f(y)=[1+f(0)]\cdot\sqrt[3]{y-1}-1.$$ Since the equation requires $x\gt0,$ it means that $y\gt1.$ However, this represents a problem, as it means that $0$ is not in the domain of $f,$ and letting $y=0$ implies $f(0)=-1,$ and we know form the original equation that $f(t)\gt-1.$
As such, the original equation has no solutions. If you replace the original integral by $$\int_a^{x^3+1},$$ where $a\gt1.$ Accordingly, this results in $$f(x)=[1+f(a)]\cdot\sqrt[3]{x-1}-1.$$ One can solve for $f(a),$ by having $$f(a)=[1+f(a)]\cdot\sqrt[3]{a-1}-1=\sqrt[3]{a-1}+\sqrt[3]{a-1}\cdot{f(a)}-1,$$ which is equivalent to $$[1-\sqrt[3]{a-1}]\cdot{f(a)}=\sqrt[3]{a-1}-1,$$ and as long as $a\neq2,$ this means $$f(a)=\frac{\sqrt[3]{a-1}-1}{1-\sqrt[3]{y-1}}=-1,$$ which means that $$f(x)=-1,$$ which is not allowed by the original equation. So in fact, the equation can only have solutions if $a=2.$ $f(a)$ then becomes a free parameter, and the family $$f(x)=[1+f(2)]\cdot\sqrt[3]{x-1}-1$$ becomes an infinite family of solutions.
