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题解
思路入口：
本题表面上是“从
\[ y=\ln \left (x+\sqrt {x^2+1}\right ) \]
中解出 \(x\)”，但若直接去解含根式方程，容易把式子平方得很繁。更好的入口是注意到
\[ \left (x+\sqrt {x^2+1}\right )\left (\sqrt {x^2+1}-x\right )=1. \]
也就是说，\(x+\sqrt {x^2+1}\) 的倒数正好是 \(\sqrt {x^2+1}-x\)。这会把对数式变成一对互补关系：
\[ e^y=x+\sqrt {x^2+1},\qquad e^{-y}=\sqrt {x^2+1}-x. \]
从思维上看，这仍然是方程组思维：把 \(x\) 和 \(\sqrt {x^2+1}\) 暂时看成两个需要消元的对象，通过两条同源方程联立，最后用相减消去根式。
详细解答：
由题设
\[ y=\ln \left (x+\sqrt {x^2+1}\right ). \]
因为 \(\sqrt {x^2+1}>|x|\)，所以
\[ x+\sqrt {x^2+1}>0, \]
对数对任意实数 \(x\) 都有意义。两边取指数，得
\[ e^y=x+\sqrt {x^2+1}. \tag {1} \]
再看 \(-y\)。由对数性质
\[ -y=-\ln \left (x+\sqrt {x^2+1}\right ) =\ln \frac {1}{x+\sqrt {x^2+1}}. \]
对分母有理化：
\[ \frac {1}{x+\sqrt {x^2+1}} =\frac {\sqrt {x^2+1}-x}{\left (\sqrt {x^2+1}+x\right )\left (\sqrt {x^2+1}-x\right )} =\sqrt {x^2+1}-x. \]
所以
\[ -y=\ln \left (\sqrt {x^2+1}-x\right ), \]
进而
\[ e^{-y}=\sqrt {x^2+1}-x. \tag {2} \]
此时 \((1),(2)\) 可以看成关于 \(x\) 与 \(\sqrt {x^2+1}\) 的二元一次方程组。用 \((1)-(2)\)，根式项抵消：
\[ e^y-e^{-y}=2x. \]
因此
\[ x=\frac {1}{2}\left (e^y-e^{-y}\right ). \]
这已经把原来的自变量 \(x\) 表示成了函数值 \(y\) 的函数，即
\[ x=f^{-1}(y)=\frac {1}{2}\left (e^y-e^{-y}\right ). \]
按通常书写习惯，将自变量记回 \(x\)，得到
\[ f^{-1}(x)=\frac {1}{2}\left (e^x-e^{-x}\right ). \]
定义域与值域：
原函数的定义域为
\[ (-\infty ,+\infty ). \]
由
\[ \eta =\frac {1}{2}\left (e^y-e^{-y}\right ) \]
可知，对任意 \(y\in \mathbb {R}\)，都能找到唯一实数 \(\eta \) 使 \(f(\eta )=y\)。所以原函数值域为 \(\mathbb {R}\)，反函数的定义域为
\[ (-\infty ,+\infty ). \]
对应地，反函数的值域等于原函数定义域，也是
\[ (-\infty ,+\infty ). \]
于是最终答案为
\[ \boxed {f^{-1}(x)=\frac {1}{2}\left (e^x-e^{-x}\right ),\qquad x\in (-\infty ,+\infty )}. \]
为什么它一定有反函数：
本题不只要会算，还要知道为什么可以谈反函数。函数
\[ f(x)=\ln \left (x+\sqrt {x^2+1}\right ) \]
是严格递增函数。可以从反函数表达式看出，\(x=\frac 12(e^y-e^{-y})\) 对每个 \(y\in \mathbb {R}\) 都给出唯一的 \(x\)；也可以在后续学到导数后验证
\[ f'(x)=\frac {1}{\sqrt {x^2+1}}>0. \]
因此原函数一一对应，有反函数。
考场满分写法：
必须写：
\[ e^y=x+\sqrt {x^2+1},\qquad e^{-y}=\sqrt {x^2+1}-x. \]
其中第二式要通过倒数有理化说明：
\[ \frac {1}{x+\sqrt {x^2+1}}=\sqrt {x^2+1}-x. \]
然后相减得
\[ x=\frac 12(e^y-e^{-y}). \]
最后交换 \(x,y\)，并写出
\[ f^{-1}(x)=\frac 12(e^x-e^{-x}),\qquad x\in \mathbb {R}. \]
可以略写的是原函数单调性的详细证明，但不能漏掉反函数定义域。
答案：
\[ \boxed {f^{-1}(x)=\frac {1}{2}\left (e^x-e^{-x}\right ),\quad x\in \mathbb {R}}. \]
知识点
核心知识点：
函数
\[ \ln \left (x+\sqrt {x^2+1}\right ) \]
常记为反双曲正弦函数 \(\operatorname {arsinh}x\)，它的反函数是双曲正弦函数
\[ \sinh x=\frac {e^x-e^{-x}}{2}. \]
本题不是要求记忆函数名，而是要掌握有理化消根号的结构：
\[ \left (x+\sqrt {x^2+1}\right )\left (\sqrt {x^2+1}-x\right )=1. \]
可迁移方法：
遇到
\[ \ln \left (x+\sqrt {x^2+a^2}\right ) \]
这类结构时，优先尝试对倒数有理化，构造 \(e^y\) 与 \(e^{-y}\)。若根式前后正负号互补，通常相加或相减可以消去根式或消去 \(x\)。
这类题的本质不是单纯“有理化技巧”，而是“制造第二个方程，再联立消元”的方程组思维。只不过第二个方程不是题目直接给的，而是由倒数有理化主动构造出来的。
母题模板：
\[ y=\ln \left (x+\sqrt {x^2+1}\right ) \quad \Longrightarrow \quad \begin {cases} e^y=x+\sqrt {x^2+1},\\ e^{-y}=\sqrt {x^2+1}-x, \end {cases} \]
所以
\[ x=\frac 12(e^y-e^{-y}). \]