How to fix this stupid mistake while keeping it as simple as it is. In this answer I used some symmetry to show that
$$
\int_0^1 \frac{dx}{x+ \sqrt{1-x^2}} = \frac \pi 4.
$$
Then I thought about whether I could make it simpler by avoiding trigonometric substitutions, like this:
$$
y = \sqrt{1-x^2}
$$
Therefore
$$
y^2+x^2=1 \tag a
$$
From the symmetry in line $(\text{a})$ we get
$$
\int_0^1 \frac{dx}{x+y} = \int_0^1 \frac{dy}{x+y}.
$$
Now let's do something stupid, claiming that the sum of these two integrals is
$$
\int_0^1 \frac{dx + dy}{x+y}.
$$
This is stupid because the expression $\displaystyle\int_0^1$ means some particular variable is going from $0$ to $1$, and surely that is not $x+y$. To write such an expression is either to hope the gullible reader won't notice the problem, or else to be the confused student who doesn't notice it. If we want to continue being stupid, we could go on to write this last integral as
$$
\int_0^1 \frac{du} u,
$$
and this is $+\infty$. (Therefore $\pi/4=+\infty/2$, we could go on to claim if we're feeling adventurous.)
How can we rescue this approach without making it complicated or lengthy? ("This approach" would have to mean fully respecting the symmetrical roles played by $x$ and $y$.)
 A: Let $A$ be the arc of the unit circle in the first quadrant,
that is, the quarter-circle arc between $(1,0)$ and $(0,1)$.
Then integrating along the arc using $ds$ to denote
the length measure, the length of the arc is
$$
\int_A ds = \frac\pi2.
$$
Now let's parameterize the path along the arc by some variable $t$
via a one-to-one continuous mapping $f$ from the interval $[0,1]$ to the
points on the arc with $f(0) = (1,0)$, so that
$$
 \int_0^1 \frac{ds}{dt} \, dt = \int_A ds = \frac\pi2.
$$
Notice that with this parameterization, $\frac{dx}{dt} \geq 0$
and $\frac{dy}{dt} \leq 0$ everywhere on the arc.
At any point along the arc,
$\frac{ds}{dt}$ is the hypotenuse of a right triangle
with legs $\frac{dx}{dt}$  and $-\frac{dy}{dt}$.
Because the direction of the path at each point $(x,y)$ is perpendicular
to the segment from $(0,0)$ to $(x,y)$,
the triangle with sides $\frac{ds}{dt}$,
$\frac{dx}{dt}$, and $-\frac{dy}{dt}$
is similar to a right triangle with hypotenuse $1$ and legs $x$ and $y$.
Hence
$$
\frac{\frac{dx}{dt} - \frac{dy}{dt}}{x + y}
 = \frac{\left(\frac{ds}{dt}\right)}{1} = \frac{ds}{dt}.
$$
As $t$ runs from $0$ to $1$, $x$ runs from $0$ to $1$ and
$y$ runs from $1$ to $0$. Therefore
\begin{align}
\int_0^1 \frac{dx}{x + y} + \int_0^1 \frac{dy}{x + y} &=
\int_0^1 \frac{dx}{x + y} - \int_1^0 \frac{dy}{x + y} \\
 &= \int_0^1 \frac{\frac{dx}{dt}\,dt}{x + y} 
    - \int_0^1 \frac{\frac{dy}{dt}\,dt}{x + y} \\
 &= \int_0^1 \frac{\frac{dx}{dt} - \frac{dy}{dt}}{x + y}\,dt \\
 &= \int_0^1 \frac{ds}{dt}\,dt \\
&= \frac\pi2.
\end{align}
Then you can use the observation that 
$$\int_0^1 \frac{dx}{x + y} = \int_0^1 \frac{dy}{x + y}$$
to conclude that $$\int_0^1 \frac{dx}{x + y} = \frac\pi4.$$
The trick here, as in the parameterization $x=\cos\theta$,
$y=\sin\theta$, is that you have to "match up" the meshes of
$dx$ and $dy$ so that you can correctly add the them together.
The trigonometric substitution was just one particular mapping
from a parameter $\theta$ to the arc.
But you don't need to invoke that particular mapping.
The mapping $x = t$, $y = \sqrt{1 - t^2}$ also works, for example.

Update: The following is an observation on notation
inspired by the comments.
The solution above is "dressed up" (or you might say "dumbed down")
to fit with my recollection of what a student would become accustomed to
in a first-year calculus course based on standard analysis.
The parameter $t$ sets a direction of integration along the arc 
and keeps the symbols $dx$, $dy$, and $ds$ from having to appear outside the usual contexts such as $\int (\text{something})\,dx$ 
or $\frac{dx}{d(\text{something})}$, 
but it really has no other reason to be defined.
Given a decent theory of differentials or infinitesimals,
so that we can speak of $ds$, $dx$, and $dy$ as objects in their own rights,
if we orient the arc so that $\int_A$ integrates from $(0,1)$ to $(1,0)$
then we have the ratios $ds : dx : -dy = 1 : y : x$ and can write
\begin{align}
\frac{dx - dy}{x + y} &= \frac{ds}{1} = ds, \\
\int_0^1 \frac{dx}{x + y} + \int_0^1 \frac{dy}{x + y}
 &= \int_0^1 \frac{dx}{x + y} - \int_1^0 \frac{dy}{x + y} \\
 &= \int_A \frac{dx}{x + y} - \int_A \frac{dy}{x + y} \\
 &= \int_A \frac{dx - dy}{x + y} \\
 &= \int_A ds \\
 &= \frac\pi2.
\end{align}
This is a more intuitive approach; or at least, I think it is, since it is
how I got the intuition on how to set up the integrals in the first place,
before I added the parameterization.

And now for a third version inspired by additional comments!
Under the orientation of $ds$ along the quarter-circle arc $A$
from $(1,0)$ to $(0,1)$, we have
$\frac{dx}{ds} = y$ and $\frac{dy}{ds} = -x$. Then
\begin{align}
\int_0^1 \frac{dx}{x + y} + \int_0^1 \frac{dy}{x + y}
 &= \int_0^1 \frac{dx}{x + y} - \int_1^0 \frac{dy}{x + y} \\
 &= \int_A \frac{dx/ds}{x + y}\,ds - \int_A \frac{dy/ds}{x + y}\,ds \\
 &= \int_A \frac{y - (-x)}{x + y}\,ds \\
 &= \int_A ds \\
 &= \frac\pi2.
\end{align}
A: I'm not sure I understand what you are after, so please don't be too hard on me if I misunderstand you.
Let
$$
I=\int_0^1 \frac{1}{x+\sqrt{1-x^2}}\,dx
$$
Changing variables via $x\mapsto \sqrt{1-x^2}$ will give you
$$
I=\int_0^1\frac{x}{\sqrt{1-x^2}(x+\sqrt{1-x^2})}\,dx.
$$
Thus
$$
\begin{aligned}
2I&=\int_0^1 \frac{1}{x+\sqrt{1-x^2}}+\frac{x}{\sqrt{1-x^2}(x+\sqrt{1-x^2})}\,dx\\
&=\int_0^1 \frac{1}{\sqrt{1-x^2}}\,dx=\arcsin 1=\frac{\pi}{2}.
\end{aligned}
$$
