Generalized Euler substitution doesn't seem to work when the integration variable has a dimension I came across Euler substitutions while trying to evaluate the integral $\int \frac{y^2}{x^2+y^2+z^2 + x\sqrt{x^2+y^2+z^2}} dy$, where $x, y, z$ are length quantities. The generalized substitution at the bottom of the page, $\sqrt{ax^2 + bx + c} = \sqrt{a} + xt$ where $x$ is the integration variable, is the one I'd prefer to memorize since it's more powerful than the first three together and has no additional conditions.
But how does it work in physics where the $x$ has a dimension? The $\sqrt{a}$ term never has the same unit as the other terms. In my case it can be written as $\sqrt{y^2 + x^2 + z^2} - 1 = yt$. I can't subtract a dimensionless quantity from a dimensional one.
The other three substitutions don't have this problem.

You can always make a substitution that makes the integration variable dimensionless, and then this isn't an issue.  Rewrite your integral as 
$$
\int \frac{y^2}{ x^2+z^2 + y^2 + x \sqrt{x^2 + z^2} \sqrt{\frac{y^2}{x^2+z^2} + 1} } dy.
$$ 
Substitute the dimensionless variable $\eta = y/\sqrt{x^2 + z^2}$:
$$
(x^2 + z^2)^{3/2} \int \frac{\eta^2 d \eta}{(x^2 + z^2)(1 + \eta^2) + x \sqrt{x^2 + z^2} \sqrt{\eta^2 + 1}} = \sqrt{x^2 + z^2} \int \frac{\eta^2 d \eta}{1 + \eta^2 + \frac{x}{\sqrt{x^2 + z^2}} \sqrt{\eta^2 + 1}}.
$$
The Euler substitution is now $\sqrt{\eta^2 + 1} = 1 + \eta t$, which is dimensionally self-consistent since both $\eta$ and $t$ are dimensionless.  In fact, the entire integral is dimensionless, and the pre-factor of $\sqrt{x^2 + z^2}$ ensures that the whole thing has dimensions of length like your original integral did. 
(Whether or not this substitution makes the integral any easier I don't know.)
