Substitute $x=\sqrt2\,\sin y$, exploit symmetry, convert from hyperbolic to logarithmic, and substitute $y=2\arctan z$:
$$\begin{align*}
I &= \int_1^{\sqrt2} \frac{\operatorname{artanh} \sqrt{2-x^2}}{1+x} \, dx \\
&= \sqrt2 \int_{\tfrac\pi4}^{\tfrac\pi2} \frac{\cos y \operatorname{artanh}\left(\sqrt2\,\cos y\right)}{1+\sqrt2\,\sin y} \, dy \\
&= \sqrt2 \int_0^{\tfrac\pi4} \frac{\cos \left(y+\frac\pi4\right) \operatorname{artanh}\left(\sqrt2\,\cos \left(y+\frac\pi4\right)\right)}{1+\sqrt2\,\sin \left(y+\frac\pi4\right)} \, d\left(y+\frac\pi4\right) \\
&= \int_0^{\tfrac\pi4} \frac{(\cos y-\sin y) \operatorname{artanh}\left(\cos y-\sin y\right)}{1+\cos y+\sin y} \, dy \tag{$*$} \\
&= \int_{\tfrac\pi2}^{\tfrac\pi4} \frac{\left(\cos\left(\frac\pi2-y\right)-\sin\left(\frac\pi2-y\right)\right) \operatorname{artanh}\left(\cos \left(\frac\pi2-y\right)-\sin \left(\frac\pi2-y\right)\right)}{1+\cos\left(\frac\pi2-y\right)+\sin \left(\frac\pi2-y\right)} \, d\left(\frac\pi2-y\right) \\
&= \int_{\tfrac\pi4}^{\tfrac\pi2} \frac{(\cos y-\sin y) \operatorname{artanh}\left(\cos y-\sin y\right)}{1+\cos y+\sin y} \, dy \tag{$**$} \\
&= \frac14 \int_0^{\tfrac\pi2} \frac{\cos y - \sin y}{1+\cos y+\sin y} \log \frac{1+\cos y-\sin y}{1-\cos y+\sin y} \, dy \tag*{$\frac{(*)+(**)}2$}\\
&= \frac14 \int_0^1 \frac{1-2z-z^2}{(1+z)\left(1+z^2\right)} \log \frac{1-z}{z(1+z)} \, dz
\end{align*}$$
Expanding into partial fractions and breaking up the logarithms leads to integrals of the much simpler forms,
$$\int_0^1 \frac{L(z)}{1+z}\,dz \text{ and } \int_0^1 \frac{2z L(z)}{1+z^2} \, dz$$
where $L(z) \in \left\{\log z, \log(1\pm z)\right\}$. Closed forms are listed below. Linked Q&As were found with Approach Zero.
$$\begin{align*}
\int_0^1 \frac{\log z}{1+z} \, dz &= -\frac{\pi^2}{12} \tag1 \\
\int_0^1 \frac{\log(1-z)}{1+z} \, dz &= \frac12 \log^22 - \frac{\pi^2}{12} \tag2 \\
\int_0^1 \frac{\log(1+z)}{1+z} \, dz &= \frac12 \log^22 \\[2ex]
\int_0^1 \frac{2z \log z}{1+z^2} \, dz &= -\frac{\pi^2}{24} \tag3 \\
\int_0^1 \frac{2z \log(1-z)}{1+z^2} \, dz &= \frac14 \log^22 - \frac{5\pi^2}{48} \tag4 \\
\int_0^1 \frac{2z \log(1+z)}{1+z^2} \, dz &= \frac14 \log^22 + \frac{\pi^2}{48} \tag4
\end{align*}$$
$(1)$, $(2)$, $(3)$, $(4)$
On the other hand, as @Zacky pointed out, by replacing $z\mapsto\dfrac{1-z}{1+z}$ one can collapse $I$ into a multiple of integral $(1)$ :
$$\begin{align*}
\int_0^1 \frac{\log \frac{1-z}{1+z}}{1+z} \, dz &= \int_0^1 \frac{\log z}{1+z} \, dz \\[2ex]
\int_0^1 \frac{2z \log \frac{1-z}{1+z}}{1+z^2} \, dz &= \int_0^1 \frac{2\log z}{1+z} \, dz - \int_0^1 \frac{2z \log z}{1+z^2} \, dz \\
&= \int_0^1 \frac{2\log z}{1+z} \, dz - \int_0^1 \frac{\frac12 \log z^2}{1+z^2} \, dz^2 \\
&= \frac32 \int_0^1 \frac{\log z}{1+z} \, dz
\end{align*}$$
In general the (signed) cross-sectional area cannot be determined from the volume of the solid of revolution it generates, that is, the definite integral $\int_a^b f \,dx$ cannot in general be determined from $\int_a^b f^2 \,dx$.
For example, the functions $f(x) := \sqrt 3 x$ and $g(x) = 1$ satisfy (over the interval $[a, b] = [0, 1]$)
$$\int_0^1 f^2 \,dx = 1 = \int_0^1 g^2 \,dx$$ but
$$\int_0^1 f \,dx = \frac{\sqrt 3}2 \qquad \textrm{and} \qquad \int_0^1 g \,dx = 1.$$
As for the integral itself,
substituting $x = e^{y^2}$ yields
$$\int_1^e \sqrt{\ln x} \,dx = 2 \int_0^1 y^2 e^{y^2} \,du ,$$
and integrating by parts with $u = y$, $dv = 2 y e^{y^2}$ gives
$$\left.y e^{y^2}\right\vert_0^1 - \int_0^1 e^{u^2} \,du = e - \int_0^1 e^{u^2} \,du.$$
The integrand $e^{u^2}$ does not have an antiderivative expressible in closed form in terms of terms of elementary functions, but in terms of the purpose-built imaginary error function,
$$\operatorname{erfi} x := -i \operatorname{erf} ix = \frac2{\sqrt \pi} \int_0^x e^{u^2} \,du,$$
we have
$$\int_1^e \sqrt{\ln x} \,dx = e - \frac{\sqrt \pi}2 \operatorname{erfi} 1 = 1.25563\ldots .$$
Best Answer
OP: the interval change still does explain why the integral cannot be solved on calculators
This is because the "dysfunction"/discrepancy that you are raising has nothing to do with the integration-by-substitution process.
If $f$ is Riemann integrable on $[a,b]$ and has antiderivative $F,$ the second fundamental theorem of calculus guarantees that $$\int_a^b f=F(b)-F(a).$$ In your example, the boldfaced condition is not satisfied (the integral fails to converge/exist due to the singularity at $x=1.5$), and $$\int_1^3 \frac{4}{(2x-3)^4} \mathrm dx\ne \left[\frac2{3 (3 - 2 x)^3}\right]_1^3,$$ even though $$\int\frac{4}{(2x-3)^4} \mathrm dx=\frac2{3 (3 - 2 x)^3}+C.$$
This, together with Martin's example in the second comment, shows that an integrand may have an antiderivative whilst not being integrable on the desired interval.