The given integral is:

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

We are tasked with evaluating this using an appropriate substitution method. A suitable substitution would involve simplifying the term inside the inverse sine function.

Step-by-Step Solution

Let's try using the substitution: $x = \tan(\theta), \quad dx = \sec^2(\theta) d\theta$ This substitution helps simplify expressions involving $1 + x^2$.

Step 1: Substituting into the integral

Using the substitution $x = \tan(\theta)$: $1 + x^2 = 1 + \tan^2(\theta) = \sec^2(\theta)$ Thus, the expression $\frac{2x}{1 + x^2}$ becomes: $\frac{2\tan(\theta)}{\sec^2(\theta)} = 2\sin(\theta)\cos(\theta)$

This simplifies the original expression for the inverse sine: $\sin^{-1}\left(2\sin(\theta)\cos(\theta)\right) = \sin^{-1}(\sin(2\theta)) = 2\theta$

Step 2: Transform the limits of integration

When $x = 0$, $\theta = 0$. When $x = 1$, $\theta = \frac{\pi}{4}$.

The new limits of integration are from $0$ to $\frac{\pi}{4}$, and the integral becomes: $\int_0^{\frac{\pi}{4}} 2\theta \cdot \sec^2(\theta) d\theta$

Step 3: Solve the transformed integral

We now have to solve: $2 \int_0^{\frac{\pi}{4}} \theta \sec^2(\theta) d\theta$

This can be solved using integration by parts. Let: $u = \theta, \quad dv = \sec^2(\theta) d\theta$ Then: $du = d\theta, \quad v = \tan(\theta)$

Now apply the integration by parts formula: $\int u dv = uv - \int v du$ Substituting: $2\left[\theta \tan(\theta) \Big|_0^{\frac{\pi}{4}} - \int_0^{\frac{\pi}{4}} \tan(\theta) d\theta\right]$

Evaluate each part:

1. $\theta \tan(\theta) \Big|_0^{\frac{\pi}{4}} = \frac{\pi}{4}$
2. $\int_0^{\frac{\pi}{4}} \tan(\theta) d\theta = \ln(\sec(\theta)) \Big|_0^{\frac{\pi}{4}} = \ln(\sec(\frac{\pi}{4})) = \ln(\sqrt{2}) = \frac{1}{2} \ln(2)$

Thus, the integral evaluates to: