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Let $x = \tan \theta$. This substitution is useful because it allows us to simplify the expression inside the $\sin^{-1}$ function using trigonometric identities.
Substitute $x = \tan \theta$ into the expression: $$ \frac{x}{\sqrt{1+x^2}} = \frac{\tan \theta}{\sqrt{1+\tan^2 \theta}} $$
Recall the trigonometric identity: $1 + \tan^2 \theta = \sec^2 \theta$. Therefore, $$ \frac{\tan \theta}{\sqrt{1+\tan^2 \theta}} = \frac{\tan \theta}{\sqrt{\sec^2 \theta}} = \frac{\tan \theta}{|\sec \theta|} $$ Assuming $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$, $\sec \theta$ is positive, so $|\sec \theta| = \sec \theta$. $$ \frac{\tan \theta}{\sec \theta} = \frac{\frac{\sin \theta}{\cos \theta}}{\frac{1}{\cos \theta}} = \sin \theta $$
Now, substitute the simplified expression back into the $\sin^{-1}$ function: $$ \sin^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right) = \sin^{-1}(\sin \theta) = \theta $$
Since we initially let $x = \tan \theta$, we have $\theta = \tan^{-1} x$. Therefore, $$ \sin^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right) = \tan^{-1} x $$
Final Answer: $\tan^{-1} x$
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