Integration and function finding

Question

I have a continuous function $f:\mathbb{R}\rightarrow\mathbb{R}$ and I want to find a function $F:\mathbb{R}\rightarrow\mathbb{R}$ which is true $F'(x)=f(x)$. Plus I have that: $$2f(x)=\frac{(x+1)^2}{x^2+1}F(x)$$ and $F(0)=1$. I am thinking of doing this: $$2F'(x)-\frac{(x+1)^2}{x^2+1}F(x)=0\iff F'(x)-\frac{(x+1)^2}{2(x^2+1)}F(x)=0$$ and then multiply by $e^{G(x)}$ with $G'(x)=g(x)$, so: $$e^{G(x)}F'(x)+e^{G(x)}g(x)F(x)=(e^{G(x)}F(x))'$$ but I don't know how to continue. Any help?

Answer 1

Let $G'(x)=-\dfrac{(x+1)^{2}}{2(x^{2}+1)}$, then \begin{align*} -\dfrac{(x+1)^{2}}{2(x^{2}+1)}&=-\dfrac{x^{2}+1+2x}{2(x^{2}+1)}\\ &=-\dfrac{1}{2}-\dfrac{x}{x^{2}+1}, \end{align*} then \begin{align*} G(x)=\int G'(x)dx=-\dfrac{1}{2}x-\dfrac{1}{2}\log(x^{2}+1), \end{align*} so $e^{G(x)}F(x)=c$ and $F(x)=ce^{-G(x)}$.