Prove symmetry of natural logarithm

Prove that $f(x)=\mathrm{ln}\sqrt{{x}^{2}+1}$ is symmetrical in $x=0$

$\mathrm{ln}\sqrt{(x-a{)}^{2}+1}=\mathrm{ln}\sqrt{(x+a{)}^{2}+1}$

$\sqrt{(x-a{)}^{2}+1}=\sqrt{(x+a{)}^{2}+1}$

$(x-a{)}^{2}+1=(x+a{)}^{2}+1$

${x}^{2}-2ax+{a}^{2}+1={x}^{2}+2ax+{a}^{2}+1$

$-2ax=2ax$

$-2ax=2ax$

$-x=x$?

I don't know what to do? Is this the proof or did I miss something?

Prove that $f(x)=\mathrm{ln}\sqrt{{x}^{2}+1}$ is symmetrical in $x=0$

$\mathrm{ln}\sqrt{(x-a{)}^{2}+1}=\mathrm{ln}\sqrt{(x+a{)}^{2}+1}$

$\sqrt{(x-a{)}^{2}+1}=\sqrt{(x+a{)}^{2}+1}$

$(x-a{)}^{2}+1=(x+a{)}^{2}+1$

${x}^{2}-2ax+{a}^{2}+1={x}^{2}+2ax+{a}^{2}+1$

$-2ax=2ax$

$-2ax=2ax$

$-x=x$?

I don't know what to do? Is this the proof or did I miss something?