Let’s evaluate the integral:

$I = \int_0^1 \sin^{-1}\left(\frac{2x}{1 + x^2}\right) \, dx$

We’ll solve this using substitution.

Step 1: Substitution

We notice that:

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

Now, let's differentiate $u$ with respect to $x$.

$\frac{du}{dx} = \frac{(1 + x^2)(2) - 2x(2x)}{(1 + x^2)^2} = \frac{2 - 2x^2}{(1 + x^2)^2} = \frac{2(1 - x^2)}{(1 + x^2)^2}$

Thus,

$du = \frac{2(1 - x^2)}{(1 + x^2)^2} \, dx$

Step 2: Adjust the Substitution

We rewrite the term $\sin^{-1}(u)$. However, this direct substitution gets slightly complicated for integration since the derivative of $u$ is complex.

Instead, observe that:

$\frac{2x}{1 + x^2} = \sin(2\theta) \quad \text{if} \quad x = \tan(\theta).$

Thus, a trigonometric substitution could be helpful. Let’s try the trigonometric substitution $x = \tan(\theta)$, which transforms the integrand.

Step 3: Trigonometric Substitution

With $x = \tan(\theta)$, we have:

$dx = \sec^2(\theta) \, d\theta, \quad 1 + x^2 = \sec^2(\theta).$

Thus,

$\frac{2x}{1 + x^2} = \sin(2\theta).$

The integral becomes:

$I = \int_0^{\pi/4} \sin^{-1}(\sin(2\theta)) \, \sec^2(\theta) \, d\theta.$

Since $\sin^{-1}(\sin(2\theta)) = 2\theta$ for $0 \leq \theta \leq \pi/4$, the integral simplifies to:

$I = \int_0^{\pi/4} 2\theta \, \sec^2(\theta) \, d\theta.$

Step 4: Evaluate the Integral

Evaluating this integral is now straightforward.