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.
 A: *

*When you substitute, you have to change the limits as well: $x$ varies between $0$ and $H$, so $u=x+h$ varies between $H$ and $x+H$.

*This is in some sense a more serious problem, and it changes the nature of the computation a lot: $u$ depends on $t$, so you can't take it out of the integral after you substitute $t=a^2+u^2$. In fact, I'm fairly sure you won't get very far with this substitution if you tried to carry on, because the expressions won't get simpler.

Here is how I would do it. Let $x+h=a\sinh{u}$. Then $dh = a\cosh{u}\, du$, and $\sqrt{a^2+(x+h)^2} = \sqrt{a^2+a^2\sinh^2{u}} = a\cosh{u}$, so we need to compute
$$ a^2 \int \cosh^2{u} \, du. $$
$\cosh^2{u} = \frac{1}{2}(1+\cosh{2u})$, which has antiderivative
$$ \frac{u}{2} + \frac{1}{4}\sinh{2u} = \frac{u}{2} + \frac{1}{2}\sinh{u}\cosh{u}. $$
Undoing the substitution then gives
$$ \int \sqrt{a^2+(x+h)^2} \, dx = \frac{1}{2} \left( a^2\arg\sinh{\left(\frac{x+h}{a}\right)} + (x+h)\sqrt{a^2+(x+h)^2} \right). $$
The first term is reexpressible using $ \arg\sinh{z}=\log{(z+\sqrt{1+z^2})}$, so
$$ \int \sqrt{a^2+(x+h)^2} \, dx = \frac{1}{2} \left( a^2\log{\left(\frac{x+h+\sqrt{a^2+(x+h)^2}}{a}\right)} + (x+h)\sqrt{a^2+(x+h)^2} \right), $$
and then you can put whatever limits in that you want.
A: Every time you make a $u$-substitution, you must change the bounds of the integral. For instance, you initially have the integral (with respect to $h$) going from $h=0$ to $h=H$. Then you make the substitution $u=x+h$. Then $u$ varies from $u=x+0=x$ to $u=x+H$. So your new integral with respect to $u$ should have lower bound $x$ and upper bound $x+H$. Similarly with the next substitution.
A: 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.
