How to find $f(x)$ if $\int_0^{x^2} (1+t)f'(t)dt=x^4$ and $\int_0^1 f(t)dt=3 $

Question

Knowing that $f:[0,+\infty)\rightarrow \mathbb{R}$ is continuous and derivable, and that:

$\int_0^{x^2} (1+t)f'(t)dt=x^4$;

$\int_0^1 f(t)dt=3 $

Determine $f(x)$.

(Note: This is supposed to be done using just Riemann integration, not differential equations)

Edit: I've tried using the first integral calculus fundamental theorem by defining $F(x)=\int_0^{x^2} (1+t)f'(t)dt$, then the fact that $f(x)$ is a primitive from $f`(x)$ to say that $f(x^2)=\int_0^{x^2} (1+t)f'(t)dt= \int_0^{x^2} f`(t) + \int_0^{x^2} tf'(t)dt=x^4$, and then deriving everything but I'm stuck there

Answer 1

By differentiating using chain rule we get $2x(1+x^{2})f'(x^{2})=4x^{3}$. This gives $f'(x^{2})=\frac {2x^{2}} {1+x^{2}}$ so $f'(t)=\frac {2t} {1+t}$. Integrating we get $f(t)=2t-2\ln (1+t)+C$. To use the fact that $\int_0^{1}f(t)dt=3$ you have to know how to integrate $\ln(1+t)$. The antiderivative of this function is $(1+t)\ln (1+t)-(1+t)$. I leave the rest to you.