# find the value of an expression ( primitives )

I have $$F:\mathbb{R}^{*}\rightarrow \mathbb{R}$$ such that $$F'(x)=\frac{1}{x}$$, for all $$x\in \mathbb{R}^{*}, F(-1)=1$$ and $$F(1)=0$$.I need to find the value of $$F(e)+F(-e)$$ (3 is the right answer)

If $$F'(x)=1/x$$ then $$f(x)=1/x$$ so $$F(x)=ln|x|$$ this mean that the result is 2 but it's wrong.How to start?

• First step: Your conclusion that $F(x) = \ln |x|$ is mistaken; all you know is that $F(x) -= \ln |x| + C$ for some constant $C$...and more important, you only know this on each domain where $F$ is continuous, i.e., the $C$ for the negative reals might be different from the one for the positive reals. – John Hughes Apr 30 '19 at 19:43

The problem with what you have done is the fact the $$\mathbb{R}^*$$ is not an interval. You have to split it into $$(0,\infty)$$ and $$(-\infty,0)$$. In both interval $$F$$ has primitives, those two primitives are not necessary the same.
If $$x>0$$ then $$F(x)=\ln x+c_1$$ since $$F(1)=0$$ then $$c_1=0$$.
If $$x<0$$, then $$-F’(x)=\frac1{-x}$$ which means $$F(x)=\ln(-x)+c_2$$. Now: $$F(-1)=c_2=1$$ thus $$F(-x)=1+\ln(-x)$$
$$F(e)=\ln e=1$$ and $$F(-e)=1+\ln(e)=2$$ therefore $$F(e)+F(-e)=1+2=3$$