Integral u-substitution of u=$\frac{2x}{1+x^2}$ I was curious about the validity of doing such a substitution in an integral:

(1)$$u=\frac{2x}{1+x^2}$$

Changing this around a bit

$$ux^2 - 2x + u = 0$$

Solving with the quadratic formula

with $a=u, b=-2, c=u$:

$$x=\frac{1 \pm \sqrt{1 - u^2}}{u}$$

I had discerned that both definitions for x (+ and -) are indeed true, plugging this into (1). However, I’m confused as to whether doing this substitution (with the intent to specifically replace parts in the integrand with x, rather than letting the differentials cancel things out) is even valid given there’s two possible definitions for x based on the original statement (1). Even if it is, I’m also not sure about how one would solve for the differentials. 

For example, in this problem:

$$\int_0^1 \frac{1}{1+x^2}\sin^{-1}\left(\frac{2x}{1+x^2}\right)dx$$
The two substitutions are both valid. Note the reciprocal relationship between the two choices

$$\frac{1 + \sqrt{1 - u^2}}{u}
=\frac{u}{1- \sqrt{1 - u^2}}$$

to see that they are not equivalent, though, with one being the inverse of the other.
The two substitutions are both valid. Note the reciprocal relationship between the two choices

$$\frac{1 + \sqrt{1 - u^2}}{u}
=\frac{u}{1- \sqrt{1 - u^2}}$$

to see that they are not equivalent, though, with one being the inverse of the other.