Asymptotic expansion of an integral as $x \to 0^{+}$ Find the asymptotic expansion of $$F(x) := \int_{x}^{1} \frac{1}{t \sqrt{1+t^2}} \ dt, \text{ as  } x \to 0^{+}$$
I tried expanding $\frac{1}{ \sqrt{1+t^2}} = 1 - \frac{t^2}{2} + \frac{3 t^{4}}{8} + \cdot \cdot \cdot$
The integral is then :$$F(x) := \int_{x}^{1} \frac{1}{t \sqrt{1+t^2}} \ dt = \bigg[\frac{1}{t} - \frac{t}{2} + \frac{3t^{3}}{8} \cdot \cdot \cdot \bigg] dt$$
I observed that the first term goes to infinity as $x \to 0^{+}$.
Can someone point out if I am going in the wrong direction?
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
&\bbox[5px,#ffd]{\left.\int_{x}^{1}{\dd t \over t\root{1 + t^{2}}}
\,\right\vert_{\ \mrm{as}\ x\ \to\ 0^{+}}}
\\[5mm] = &\
\int_{x}^{1}{\dd t \over t} + \int_{x}^{1}
\pars{{1 \over t\root{1 + t^{2}}} - {1 \over t}}\dd t
\\[5mm] = &\
-\ln\pars{x} +
\int_{x}^{1}{1 - \root{1 + t^{2}} \over t\root{1 + t^{2}}}\,\dd t
\\[5mm] = &\
-\ln\pars{x} - \int_{x}^{1}
{t \over 1 + t^{2} + \root{1 + t^{2}}}\,\dd t
\\[5mm] = &\
-\ln\pars{x} - {1 \over 2}\int_{0}^{1}
{\dd t \over 1 + t + \root{1 + t}}
\\[2mm] &\ +
{1 \over 2}\int_{0}^{x^{2}}
{\dd t \over 1 + t + \root{1 + t}}
\\[5mm] \stackrel{\mrm{as}\ x\ \to\ 0^{+}}{\sim}\,\,\, &\
-\ln\pars{x} - {1 \over 2}\int_{0}^{1}
{\dd t \over 1 + t + \root{1 + t}}\,\dd t
\\[2mm] & \phantom{A} +
{1 \over 2}\int_{0}^{x^{2}}
\pars{{1 \over 2} - {3t \over 8} + {5t^{2} \over 16}}\dd t
\\[5mm] = &\
-\ln\pars{x}\ -\ \underbrace{{1 \over 2}\int_{0}^{1}
{\dd t \over 1 + t + \root{1 + t}}\,\dd t}
_{\mbox{a constant}}
\\[2mm] &\ +
{1 \over 4}\,x^{2} - {3 \over 32}\,x^{4} + {5 \over 96}\,x^{6}
\end{align}
A: HINT
I would start by making the substitution $t = \sinh(u)$. Thus we get
\begin{align*}
\int\frac{\mathrm{d}t}{t\sqrt{1+t^{2}}} = \int\frac{\cosh(u)}{\sinh(u)\cosh(u)}\mathrm{d}u = \int\frac{\mathrm{d}u}{\sinh(u)} = \int\frac{\sinh(u)}{\sinh^{2}(u)}\mathrm{d}u = \int\frac{\sinh(u)}{\cosh^{2}(u)-1}\mathrm{d}u
\end{align*}
Now you can make the substitution $v = \cosh(u)$, which leads to
\begin{align*}
\int\frac{\sinh(u)}{\cosh^{2}(u) - 1}\mathrm{d}u = \int\frac{\mathrm{d}v}{v^{2} - 1} = \frac{1}{2}\int\left(\frac{1}{v-1} - \frac{1}{v+1}\right)\mathrm{d}v
\end{align*}
Can you take it from here?
