= Solution
In the <integral> from part (d), set $y=\sinh^2u$. Its upper endpoint is $u_x=\operatorname{arsinh}\sqrt x$, equal to $\theta/2$ in part (b). The <hyperbolic function> identity
$$
\left(\frac{y}{1+y}\right)^{3/2}dy
=2\frac{\sinh^4u}{\cosh^2u}\,du
=2(\cosh^2u-2+\operatorname{sech}^2u)\,du
$$
gives the <antiderivative>
$$
I(x)=\left[\frac12\sinh(2u)-3u+2\tanh u\right]_0^{u_x}
=\sqrt{x(1+x)}+2\sqrt{\frac{x}{1+x}}-3\operatorname{arsinh}\sqrt x.
$$
Multiplication by $\delta_1$ therefore yields
$$
\boxed{\delta_2(x)=1+\frac3x-3\sqrt{\frac{1+x}{x^3}}\operatorname{arsinh}\sqrt x
=1+\frac3x-3\sqrt{\frac{1+x}{x^3}}\ln(\sqrt x+\sqrt{1+x}).}
$$
At small $x$, cancellation between the terms leaves the <Taylor series> $\delta_2=2x/5-8x^2/35+16x^3/105+O(x^4)$. At large $x$, $\operatorname{arsinh}\sqrt x=\tfrac12\ln(4x)+O(x^{-1})$, so the correction to one is $[3-\tfrac32\ln(4x)]/x+O(\ln x/x^2)$. Both checks agree with part (d); the <integral> or the small-$x$ <Taylor series> is preferable to subtracting large nearly equal terms in a numerical evaluation near zero.
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