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?

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?
