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I have a function $f:\mathbb{R}\rightarrow\mathbb{R}$ which is continuous at $\mathbb{R}$ and can derives at $\mathbb{R}^*$. Plus I have that $f(1)=1$ and $f'(1)=1$. Now I have: $$\frac{f(x)}{x^2}=\begin{cases}\ln{|x|}+\frac{1}{x^2}+c_1,x<0\\\ln{|x|}+\frac{1}{x^2}+c_2,x>0\end{cases}$$ and I want to find $c_1$ and $c_2$. I can easily find $c_2$, but how can I find $c_1$?

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  • $\begingroup$ You definitely want $c_1 = c_2$, that's easy to see. $\endgroup$ – Matti P. Apr 26 '18 at 10:30
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With an arbitrary $c_1$ the function $f$ is continuous.

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  • $\begingroup$ I want to prove that $c_1=c_1$ $\endgroup$ – Leos Kotrop Apr 26 '18 at 10:38
  • $\begingroup$ Prove $c_1=c_1$? :-O Sorry, but is it not always true? :) $\endgroup$ – Logic_Problem_42 Apr 26 '18 at 11:23

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