Showing this integral is always greater than $\pi/2$ Define $$I(b)= \int_1^\infty {(u+b)^{1/2} du\over (u^2+b)}$$
I want to show that $\mathit I(b)$ is greater than $\pi/2$, for every $\mathit b >0. $
 A: The following is a more-or-less elementary approach:
Changing variables $u = bv$ in the original $I(b)$ integral, we get
$$ I(b) = \int_{1/b}^{\infty} \frac{b^{1/2} \sqrt{v+1}}{(b^{1/2}v)^2 +1} \, dv = \int_{1/b}^{\infty} \sqrt{v+1} \, \partial_v(\arctan(b^{1/2}v)-\pi/2) \, dv.$$
Integrating by parts, we get
$$I(b) = \sqrt{1/b + 1} (\pi/2 - \arctan(1/b^{1/2})) + \int_{1/b}^{\infty}\frac{(\pi/2 - \arctan(b^{1/2}v)}{2\sqrt{v+1}} \, dv.$$
As mentioned before, $\limsup_{b \to \infty} I(b) \ge \pi/2,$ which also follows from this last identity. In order to prove the result, it suffices to show that $I(b)$ is nonincreasing in $b>0.$ In order to do so, it suffices to prove that the function
$$g(t) = \sqrt{\frac{1}{t^2} + 1} (\pi/2 - \arctan(1/t)) + \int_{1/t^2}^{\infty} \frac{\pi/2 - \arctan(tv)}{2\sqrt{v+1}} \, dv $$
is nonincreasing. Before we do that, we remark that a function
$$\eta(t) = \int_{a(t)}^{\infty} F(x,t) \, dx$$
is, given $a,F$ are smooth, is differentiable, and its derivative is
$$\eta'(t) = -a'(t) F(a(t),t) + \int_{a(t)}^{\infty} \partial_t F(x,t) \, dx.$$
The derivative of $\int_{1/t^2}^{\infty} \frac{\pi/2 - \arctan(tx)}{2\sqrt{x+1}} \, dx$ is then
$$ 2 t^{-3} \frac{\pi/2 - \arctan(1/t)}{2\sqrt{\frac{1}{t^2} + 1}} - \int_{1/t^2}^{\infty} \frac{x}{2\sqrt{x+1}} \frac{1}{(xt)^2+1} \, dx,$$
while that of $\sqrt{\frac{1}{t^2}+1}(\pi/2 - \arctan(1/t))$ is
$$ -2 t^{-3} \frac{1}{2\sqrt{\frac{1}{t^2} + 1}} (\pi/2 - \arctan(1/t)) + \frac{1}{t(1+t^2)^{1/2}}.$$
Therefore,
$$g'(t) = \frac{1}{t} \frac{1}{(1+t^2)^{1/2}}  -\int_{1/t^2}^{\infty} \frac{x}{2\sqrt{x+1}} \frac{1}{(xt)^2+1} \, dx. $$
Change once more variables $x = v/t^2$ in the integral on the right. This yields
$$g'(t) = \frac{1}{t(1+t^2)^{1/2}} - \frac{1}{t} \int_1^{\infty} \frac{v}{2\sqrt{v+t^2}} \frac{1}{v^2 + t^2} \, dv.$$
We then use the Cauchy-Schwarz inequality $\sqrt{v+t^2} \le (v^2 + t^2)^{1/4} (1+t^2)^{1/4}$ in the integral, which yields
$$g'(t) \le \frac{1}{t} \left( \frac{1}{(1+t^2)^{1/2}} -\frac{1}{(1+t^2)^{1/4}} \int_1^{\infty} \frac{v}{2(v^2+t^2)^{5/4}} \right) \, dv.$$
But then $-\frac{v}{2(v^2 + t^2)^{5/4}} = \partial_v ((v^2+t^2)^{-1/4}),$ so the integral evaluates explicitly to $\frac{1}{(1+t^2)^{1/4}}.$ Thus $g'(t) \le 0.$ Notice, moreover, that the inequality is strict (due to the Cauchy-Schwartz inequality) unless $t=0.$ Therefore, $g$ is strictly decreasing on $t>0,$ and so is $I(b),$ finishing the problem.
A: $$J=\int\frac{\sqrt{u+b}}{u^2+b}\,du$$ Let $u=t^2-b$ to make
$$J=\int\frac{2 t^2}{t^4-2 b t^2+b(b+1)} \,dt=\int\frac{2 t^2}{\left(t^2-r\right) \left(t^2-s\right)}\,dt$$ Partial fraction decomposition
$$J=\frac 2{r-s}\int \left(\frac{r}{t^2-r}-\frac{s}{t^2-s}\right)\,dt=\frac 2{r-s}\left(\sqrt{s} \tanh ^{-1}\left(\frac{t}{\sqrt{s}}\right)-\sqrt{r} \tanh
   ^{-1}\left(\frac{t}{\sqrt{r}}\right)\right)$$ Now, back to $u$, replace $r$ and $s$, use the bounds. Now, Taylor expansion for large $b$ gives
$$I(b)=\frac{\pi }{2}+\frac{\log (b)+4 \log (2)-2}{4\sqrt b}+\frac{\pi }{16 b}+O\left(\frac{1}{b^{3/2}}\right)$$
Trying for $b=100$, the exact result is $1.70710$ while the above truncated expansion gives $1.70720$.
Similarly, for small values of $b$
$$I(b)=2-\frac{b}{15}+\frac{37 b^2}{1260}+O\left(b^{3}\right)$$
Trying for $b=1$, the exact result is $1.95222$ while the above truncated expansion gives $1.96270$.
