Step 1: Substitute $x=\sin\theta$:
Let $x=\sin\theta$, so $dx=\cos\theta\,d\theta$ and $\sqrt{1-x^2}=\cos\theta$.
Also $\sin^{-1}x=\theta$.
Step 2: Rewrite the integral in terms of $\theta$:
\[ \int\frac{x\sin^{-1}x}{\sqrt{1-x^2}}dx=\int\frac{\sin\theta\cdot\theta}{\cos\theta}\cos\theta\,d\theta=\int\theta\sin\theta\,d\theta \]
The square root and the cosine from dx cancel exactly, leaving a simple product.
Step 3: Integrate by parts in $\theta$:
Take $p=\theta$, $dq=\sin\theta\,d\theta$, so $dp=d\theta$, $q=-\cos\theta$.
\[ \int\theta\sin\theta\,d\theta=-\theta\cos\theta+\int\cos\theta\,d\theta=-\theta\cos\theta+\sin\theta+C \]
Step 4: Return to the variable x:
Replace $\theta=\sin^{-1}x$, $\cos\theta=\sqrt{1-x^2}$, and $\sin\theta=x$.
\[ -\theta\cos\theta+\sin\theta+C=-\sin^{-1}x\cdot\sqrt{1-x^2}+x+C \]
Final Answer:
This matches the result from direct integration by parts.
\[ \boxed{\int\frac{x\sin^{-1}x}{\sqrt{1-x^2}}dx=x-\sqrt{1-x^2}\sin^{-1}x+C} \]