Clarification Needed Regarding $\sinh^{-1}(-3)$

Question

As the definition of $\sinh^{-1}(x)$ goes :

$\sinh^{-1}(x)=\ln\left(x+\sqrt{x^{2}+1}\right)$

So what I expect to get is

$\sinh^{-1}(-3)=\ln\left(-3+\sqrt{10}\right)$

The value inside of the natural logarithm is positive because I can estimate that $\sqrt{10}$ is bigger than 3

However the answer in my textbook is $-\ln\left(3+\sqrt{10}\right)$

They claim that $\ln\left(-3+\sqrt{10}\right)=-\ln\left(3+\sqrt{10}\right)$

I know that $\ln(\frac{1}{x})=-\ln(x)$ But I am unable to make the connection with the answer in my textbook, is there any explanation for this ? is my first answer is correct ?

Answer 1

$$ \sqrt{10}-3=\frac{(\sqrt{10}-3)(\sqrt{10}+3)}{\sqrt{10}+3}=\frac{1}{\sqrt{10}+3}. $$

Answer 2

Note that $$-3+\sqrt{10}=(-3+\sqrt{10})\cdot\frac{-3-\sqrt{10}}{-3-\sqrt{10}}=\frac{-1}{-3-\sqrt{10}}=(3+\sqrt{10})^{-1}$$