Show that y=sqrt(x²+1)- ln (sqrt(1+x²)-1)/x (the log's

Show that $y=\sqrt{x^2+1}-\ln\dfrac{\sqrt{1+x^2}-1}{x}$ (the log's argument taken positive for $x>0$) can be written $y=ax+b+\dfrac cx+o\!\left(\dfrac1x\right)$ as $x\to+\infty$, and determine $a,b,c$.

Step 1: Substitute $t=1/x\to0^+$ in the logarithm

\[ \frac{\sqrt{1+x^2}-1}x = \frac{\sqrt{1+t^2}/t-1}{1/t} = \sqrt{1+t^2}-t = 1-t+\frac{t^2}2+O(t^4). \tag{1} \]

Step 2: Expand the logarithm to order $t^3$

Writing $\sqrt{1+t^2}-t=1+v$ with $v=-t+\tfrac{t^2}2+O(t^4)$, and using $\ln(1+v)=v-\tfrac{v^2}2+\tfrac{v^3}3-\cdots$: \[\ln(1+v) = \left(-t+\frac{t^2}2\right) - \left(\frac{t^2}2-\frac{t^3}2\right) + \left(-\frac{t^3}3\right) + O(t^4) = -t+\frac{t^3}6+O(t^4). \tag{2}\]

Reverting to $x=1/t$: \[\ln\frac{\sqrt{1+x^2}-1}x = -\frac1x+\frac1{6x^3}+O\!\left(\frac1{x^4}\right). \tag{3}\]

Step 3: Expand $\sqrt{x^2+1}$ for large $x$

\[ \sqrt{x^2+1} = x\sqrt{1+\frac1{x^2}} = x+\frac1{2x}+O\!\left(\frac1{x^3}\right). \tag{4} \]

Step 4: Form $y$ and collect

\[ y = \left(x+\frac1{2x}+O(x^{-3})\right) - \left(-\frac1x+\frac1{6x^3}+O(x^{-4})\right) = x+\frac3{2x}+O\!\left(\frac1{x^3}\right). \tag{5} \]

Step 5: Verification

\[x=1000\text{: }y\approx1000.0015000,\ x+\tfrac3{2x}=1000.0015,\ |{\rm error}|\approx2.9\times10^{-10}\ \checkmark\]