= Solution
Put $x=r/r_1$. The <NFW profile> becomes $\rho_D=\rho_1/[x(1+x)^2]$. Integrating spherical shells gives
$$
\begin{aligned}
M_D(r)&=4\pi\rho_1r_1^3\int_0^x\frac{u}{(1+u)^2}\,du\\
&=4\pi\rho_1r_1^3\left[\log(1+x)-\frac{x}{1+x}\right].
\end{aligned}
$$
Define $F(x)=[\log(1+x)-x/(1+x)]/x$. The spherical <circular speed> is therefore
$$
\boxed{v_{\rm NFW,c}^2(r)=\frac{GM_D(r)}r=4\pi G\rho_1r_1^2F(x).}
$$
At small $x$, the bracket is $x^2/2-2x^3/3+O(x^4)$, giving
$$
v_{\rm NFW,c}^2\simeq2\pi G\rho_1r_1r,
\qquad v_{\rm NFW,c}\propto r^{1/2}.
$$
At large $x$, the bracket is $\log x-1+O(1/x)$, giving
$$
v_{\rm NFW,c}\simeq\sqrt{4\pi G\rho_1r_1^2}\,
\left(\frac{\log x-1}{x}\right)^{1/2}.
$$
Thus the curve rises from zero, reaches one broad maximum, and eventually falls as $[\log r/r]^{1/2}$. The untruncated <NFW profile> has logarithmically divergent mass, but its circular speed still tends to zero.
Differentiating $F$ shows that its stationary point satisfies
$$
\frac{x^2}{(1+x)^2}=\log(1+x)-\frac{x}{1+x}.
$$
The numerator of $F'$ has derivative $x(1-x)/(1+x)^3$: it increases up to one and decreases thereafter, crossing zero once at positive $x>1$. Numerical bisection gives $x_{\max}\simeq2.16258$ and $F(x_{\max})\simeq0.2162166$. Hence
$$
\boxed{r_{\max}\simeq2.163r_1,\qquad
v_{\rm NFW,c,max}\simeq1.648\,r_1\sqrt{G\rho_1}.}
$$
Equivalently $v_{\max}\simeq0.4650\sqrt{4\pi G\rho_1r_1^2}$. This is the <NFW mass and circular-speed profile>.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2002/iii/paper-44-nfw-rotation.png]
{title=NFW circular speed, its maximum and inner and outer asymptotic scalings}
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