Step 1: Simplify LHS:
Using logarithmic definitions:
\( \sinh^{-1} x = \ln(x + \sqrt{x^2+1}) \implies \sinh^{-1} 2 = \ln(2 + \sqrt{5}) \).
\( \cosh^{-1} x = \ln(x + \sqrt{x^2-1}) \implies \cosh^{-1} \sqrt{6} = \ln(\sqrt{6} + \sqrt{5}) \).
Sum \( S = \ln((2+\sqrt{5})(\sqrt{6}+\sqrt{5})) \).
LHS \( = e^S = (2+\sqrt{5})(\sqrt{6}+\sqrt{5}) \).
Expand:
\[ \text{LHS} = 2\sqrt{6} + 2\sqrt{5} + \sqrt{30} + 5 \]
\[ = 5 + 2\sqrt{5} + 2\sqrt{6} + \sqrt{30} \]
Step 2: Analyze RHS Structure:
RHS \( = a + (b+\sqrt{c})\sqrt{a} + b\sqrt{c} \)
Expand:
\[ \text{RHS} = a + b\sqrt{a} + \sqrt{ac} + b\sqrt{c} \]
Step 3: Compare Coefficients:
Equate LHS and RHS:
\[ 5 + 2\sqrt{5} + 2\sqrt{6} + \sqrt{30} = a + b\sqrt{a} + b\sqrt{c} + \sqrt{ac} \]
By inspection:
Integer part: \( a = 5 \).
Term with \( \sqrt{a} = \sqrt{5} \): \( b\sqrt{5} = 2\sqrt{5} \implies b = 2 \).
Term with \( b\sqrt{c} = 2\sqrt{c} \): Matches \( 2\sqrt{6} \implies c = 6 \).
Check the last term \( \sqrt{ac} \): \( \sqrt{5 \times 6} = \sqrt{30} \). Matches perfectly.
Values are \( a=5, b=2, c=6 \).
We need \( a+b+c = 5 + 2 + 6 = 13 \).
Step 4: Required Answer:
The sum is 13.