Is this the proper use of U-Substitution? I am part of an R&D role at my job, and I have been given the task to model the magnetic field of a bowl shaped permanent magnet. So this isn't part of a textbook and I have no idea if the answer I am getting is actually correct for now, therefore I am more interested to make sure my procedure is correct of using the U-Substitution. I am essentially trying to find the total sum of the the distances of the surface of a bowl to a point "P" somewhere on the +y axis.
This is my integral:
$$
r = \int_0^H \sqrt{a^2+ (x+h)^2} \space\space dh
$$
These are my steps:
let $u =x +h$ ;$\frac{du}{dh}=1$
$$
r =\int_0^H \sqrt{a^2+ u^2} \space \space du
$$
let $t = a^2+u^2 $ ; $\frac{dt}{du}=2u$
$$
r =\frac{1}{2u} \int_0^H \sqrt{t} \space \space dt
$$
Solving and plugging back in:
$$
r =\frac{1}{2u} \left(\frac{2}{3} t ^ {\frac{3}{2}}\right)
$$
Plugging all back in:
$$
\frac{1}{2(x+h)} \left(\frac{2}{3}[a^2+(x+h)^2]^\frac{3}{2} \right)
$$
Simplifying...
$$
r = \frac{(a^2 + [x+h]^2)^\frac{3}{2}}{2(x+h)}
$$
There you have it? Are these steps correct?
Thank you very much for helping out.

I know this question has been resolved but here's another way using integration by parts and an integration formula which, in my experience, has been pretty useful to know.  I'm posting this because even though it's a "longer" route, I do think it's good to see alternative approaches.  And to me it feels more straightforward or natural, especially since I have no experience using hyperbolic trig substitutions.
$$
r = \int_0^H \sqrt{a^2+ (x+h)^2} \, dh
$$
First substitute $t = x+h$, so then $dt = dh$ and the new limits of integration are $x$ and $x+H$.
$$
r = \int_x^{x+H} \sqrt{a^2+ t^2} \, dt
$$
Now integrate by parts.  Let $u = \sqrt{a^2+t^2}$, so then $dv = dt$, $v = t$, and $du = \dfrac t{\sqrt{a^2+t^2}} \, dt$ and we have
\begin{align*}
  r &= \int_x^{x+H} \sqrt{a^2+ t^2} \, dt\\[0.3cm]
    &= t\sqrt{a^2+t^2}\bigg|_{t=x}^{t=x+H} - \int_x^{x+H} \frac{t^2}{\sqrt{a^2+t^2}} \, dt\\[0.3cm]
    &= (x+H)\sqrt{a^2+(x+H)^2} - x\sqrt{a^2+x^2} - \int_x^{x+H} \frac{\color{red}{a^2} +t^2\color{red}{-a^2}}{\sqrt{a^2+t^2}} \, dt\\[0.3cm]
    &= (x+H)\sqrt{a^2+(x+H)^2} - x\sqrt{a^2+x^2} - \int_x^{x+H} \frac{a^2 +t^2}{\sqrt{a^2+t^2}} \, dt + \int_x^{x+H} \frac{a^2}{\sqrt{a^2+t^2}} \, dt\\[0.3cm]
    &= (x+H)\sqrt{a^2+(x+H)^2} - x\sqrt{a^2+x^2} - \underbrace{\int_x^{x+H} \sqrt{a^2+t^2} \, dt}_{\text{This is $r$}} + a^2 \ln(\sqrt{a^2+t^2}+t)\bigg|_{t=x}^{t=x+H}\\[0.3cm]
   r &= (x+H)\sqrt{a^2+(x+H)^2} - x\sqrt{a^2+x^2} - r + a^2 \ln(\sqrt{a^2+(x+H)^2}+x+H) - a^2\ln(\sqrt{a^2+x^2}+x)
\end{align*}
Now add $r$ to both sides and divide by 2.
