I am asked to find the first two terms in the of the solution to the integral
$$I = \int_0^\infty \frac{rx}{(r^2+x)^{3/2}(1+x)}\,{\rm d}x$$
as $r\to0$. So firstly,
\begin{align} I &= r\int_0^\infty \frac{1+x-1}{(r^2+x)^{3/2}(1+x)}\,{\rm d}x \\ &=r\int_0^\infty \frac{1}{(r^2+x)^{3/2}}\,{\rm d}x-r\int_0^\infty\frac{1}{(r^2+x)^{3/2}(1+x)}\,{\rm d}x \\ &=2-r\int_0^\infty\frac{1}{(r^2+x)^{3/2}(1+x)}\,{\rm d}x =2-rI_1\\ \end{align}
Now let $\delta\gg r^2$ and $\delta\ll 1$ so that
$$I_1 = \int_0^\delta\frac{1}{(r^2+x)^{3/2}(1+x)}\,{\rm d}x+\int_\delta^\infty \frac{1}{(r^2+x)^{3/2}(1+x)}\,{\rm d}x$$
Note that in the first integral $x\le\delta\ll1$ so we may expand $(1+x)^{-1}$ and in the second, $x\ge\delta\gg r^2$ so we may expand $(r^2+x)^{-3/2}=x^{-3/2}(1+r^2x^{-1})^{-3/2}$. This gives
\begin{align} I_1 &= \int_0^\delta\frac{1}{(r^2+x)^{3/2}}(1-x+\ldots)\,{\rm d}x+\int_\delta^\infty \frac{1}{x^{3/2}(1+x)}\left(1-\frac{3r^2}{2x}+\ldots\right)\,{\rm d}x \\ &= \int_0^\delta\frac{1}{(r^2+x)^{3/2}}\,{\rm d}x-\int_0^\delta\frac{x}{(r^2+x)^{3/2}}\,{\rm d}x+\int_\delta^\infty \frac{1}{x^{3/2}(1+x)}\,{\rm d}x-\frac{3r^2}2\int_\delta^\infty \frac{1}{x^{5/2}(1+x)}\,{\rm d}x\\ &= J_1-J_2+J_3-\frac{3r^2}2J_4 \end{align}
All of which are integrable and (skipping the working) integrate to
$$J_1=\left[-\frac{2}{\sqrt{x+r^2}}\right]_0^\delta=\frac2r-\frac{2}{\sqrt{\delta+r^2}}$$
$$J_2=\left[\frac{2(x+2r^2)}{\sqrt{x+r^2}}\right]_0^\delta=\frac{2(\delta+2r^2)}{\sqrt{\delta+r^2}}-4r$$
$$J_3=\left[-\frac{2}{\sqrt{x}}-2\arctan{(\sqrt{x})}\right]_\delta^\infty = -\pi+\frac2{\sqrt{\delta}}+2\arctan{(\sqrt{\delta})}$$
$$J_4=\left[\frac{6x-2}{x^{3/2}}+2\arctan{(\sqrt{x})}\right]_\delta^\infty = \pi-\frac{6\delta-2}{\delta^{3/2}}-2\arctan{(\sqrt{\delta})}$$
So we have that
\begin{align} I_1&=\frac2r-\frac{2}{\sqrt{\delta+r^2}}-\frac{2(\delta+2r^2)}{\sqrt{\delta+r^2}}+4r-\pi+\frac2{\sqrt{\delta}}+2\arctan{(\sqrt{\delta})}-\frac{3r^2\pi}2-\frac{3r^2(3\delta-1)}{\delta^{3/2}}-3r^2\arctan{(\sqrt{\delta})}\\ &=-\frac{2}{\sqrt{\delta^2+r}}(2r^2+\delta+1)+\frac2r(2r^2+1)-\frac\pi2(3r^2+2)+(2-3r^2)\arctan{(\sqrt{\delta})}+\delta^{-3/2}(2\delta-3r^2(3\delta-1)) \end{align}
and finally,
$$I=2+\frac{2r}{\sqrt{\delta^2+r}}(2r^2+\delta+1)-2(2r^2+1)+\frac\pi2r(3r^2+2)-r(2-3r^2)\arctan{(\sqrt{\delta})}-r\delta^{-3/2}(2\delta-3r^2(3\delta-1))$$
However, now I'm stuck because we don't want $\delta$'s in the final solution (since they weren't included in the question). How do I get rid of them?
I do not know if this is an answer since I do not follow your path.
In all the following, I shall assume $r>0$.
Considering the integrand in $I_1$ $$\frac{1}{(r^2+x)^{3/2}(1+x)}=\frac{1}{(r^2+x)^{1/2}}\left(\frac{1}{(r^2+x)(1+x)} \right)$$ Now using partial fraction decomposition $$\frac{1}{(r^2+x)(1+x)}=\frac 1{1-r^2}\left(\frac{1}{ r^2+x}-\frac{1}{ x+1}\right)$$ $$\frac{1}{(r^2+x)^{3/2}(1+x)}=\frac 1{1-r^2}\left(\frac{1}{ (r^2+x)^{3/2}}-\frac{1}{ (x+1)(r^2+x)^{1/2}}\right)$$ $$(1-r^2)I_1=\int_0^\infty\frac{dx}{ (r^2+x)^{3/2}}-\int_0^\infty \frac{dx}{ (x+1)(r^2+x)^{1/2}}=\frac 2r-\int_0^\infty \frac{dx}{ (x+1)(r^2+x)^{1/2}}$$ Now, the unpleasant part (I shall let it to you) $$J=\int \frac{dx}{ (x+1)(r^2+x)^{1/2}}=\frac{2 }{\sqrt{1-r^2}}\tan ^{-1}\left(\sqrt{\frac{r^2+x}{r^2+1}}\right)$$ which makes $$K=\int_0^\infty \frac{dx}{ (x+1)(r^2+x)^{1/2}}=\frac{\pi -2 \sin ^{-1}(r)}{\sqrt{1-r^2}}$$ All of this makes $$(1-r^2)I_1=\frac 2r-\frac{\pi -2 \sin ^{-1}(r)}{\sqrt{1-r^2}}$$ $$I=2-\frac r{{1-r^2}}\left(\frac 2r-\frac{\pi -2 \sin ^{-1}(r)}{\sqrt{1-r^2}}\right)$$ Now, use Taylor series around $r=0$.
Being sure that you will properly finish, let me show how good is the approximation using the first two terms $$\left( \begin{array}{ccc} \text{r} & \text{exact} & \text{approx} \\ 0.1 & 0.278391 & 0.274159 \\ 0.01 & 0.0310206 & 0.0310159 \\ 0.001 & 0.00313760 & 0.00313759 \\ 0.0001 & 0.000314119 & 0.000314119 \\ 0.00001 & 0.0000314155 & 0.0000314155 \end{array} \right)$$
Edit
I did not have enough time on yesterday to add a few comments :