Re-arranging/solving $\frac{1}{a}=\int_{0}^{1}\frac{\tanh(\frac{1}{b}(x^{2}+\Delta^{2})^{1/2})}{(x^{2}+\Delta^{2})^{1/2}}dx$ , for $\Delta$. My question is a simple one: Is it possible to re-arrange the following equation for $\Delta$? Or make $\Delta$ the subject of this equation. Or, put $\Delta$ on the left hand side of the equation. Whatever's your syntax for this :-). 
$$\frac{1}{a}=\int_{0}^{1}\frac{\tanh(\frac{1}{b}(x^{2}+\Delta^{2})^{1/2})}{(x^{2}+\Delta^{2})^{1/2}}dx$$
In this equation, $a$ and $b$ are constants.
 A: I do not think that there is an explicit expression of $\Delta$ in terms of $a$. However something can be done in certain cases.
Let
$$
F(\Delta,b)=\int_{0}^{1}\frac{\tanh\Bigl(\dfrac{\sqrt{x^{2}+\Delta^{2}}}{b}\Bigr)}{\sqrt{x^{2}+\Delta^{2}}}\,dx,\quad\Delta>0,\quad b>0.
$$
It is easy to see that for any $b>0$, $F(\Delta,b)$ is decreasing as a function of $\Delta$ and $\lim_{\Delta\to\infty}F(\Delta,b)=0$. Thus, if $a>1/F(0,b)$, then the equation $F(\Delta,b)=1/a$ has no solution, while if $0<a\le1/F(0,b)$ then there exists a unique solution.
Define now
$$\begin{align*}
f(\Delta,b)&=\tanh\Bigl(\frac{\Delta}{b}\Bigr)\log\Bigl(\frac{1+\sqrt{1+\Delta^2}}{\Delta}\Bigr),\\
g(\Delta,b)&=\min\Bigl\{F(0,b),\tanh\Bigl(\frac{\sqrt{1+\Delta^2}}{b}\Bigr)\log\Bigl(\frac{1+\sqrt{1+\Delta^2}}{\Delta}\Bigr)\Bigr\}.
\end{align*}$$
Then
$$
f(\Delta,b)\le F(\Delta,b)\le g(\Delta,b).
$$
In the following picture, the lines in red are the graph of $f$ and $g$, and the blue line the graph of $F$ for $b=2$.

If $a$ is large compared to $1/F(0,b)$, then the solution of $F(\Delta,b)=1/a$ will be close to the solutions of $f(\Delta,b)=1/a$ and $g(\Delta,b)=1/a$. These equations can be solved numerically. Again if $a$ is large, an approximation is $\Delta\approx a$, or even better
$$
\Delta\approx\frac{2\,e^{1/a}}{e^{2/a}-1}.
$$
A: In short, no.
What you can do however, is note that your integrand can be very well approximated by a quadratic in $x$:
$$\frac{\tanh
   \left(\frac{\sqrt{\Delta^2+x^2}}{b}\right)}{\sqrt{\Delta^2+x^2}} \sim x^2 \frac{ \left(\Delta\ \text{sech}^2\left(\frac{\Delta}{b}\right)-b \tanh
   \left(\frac{\Delta}{b}\right)\right)}{2 b \Delta^3}+\frac{\tanh
   \left(\frac{\Delta}{b}\right)}{\Delta}$$
Then you integrate, and you end up with another equation with no closed form solution. so you'll need to use yet another approximation, but this depends on the values of $b$ and $d$.
