`int_((x sin^(-1)x)/(sqrt(1-x)^(2)))dx`

Answer

To solve the integral

\[ I = \int \frac{x \sin^{-1} x}{\sqrt{1 - x^2}} \, dx, \]

we will use the substitution method and integration by parts.

### Step 1: Rewrite the Integral
We start by rewriting the integral in a clearer form:

\[ I = \int \frac{x \sin^{-1} x}{\sqrt{1 - x^2}} \, dx. \]

### Step 2: Substitution
Let us use the substitution \( t = \sin^{-1} x \). Then, we differentiate:

\[ dt = \frac{1}{\sqrt{1 - x^2}} \, dx \quad \Rightarrow \quad dx = \sqrt{1 - x^2} \, dt. \]

Also, since \( x = \sin t \), we can substitute \( x \) in terms of \( t \):

\[ I = \int \sin t \cdot t \, dt. \]

### Step 3: Integration by Parts
Now we will apply integration by parts. We choose:

- \( u = t \) (hence \( du = dt \))
- \( dv = \sin t \, dt \) (hence \( v = -\cos t \))

Using the integration by parts formula:

\[ \int u \, dv = uv - \int v \, du, \]

we get:

\[ I = t (-\cos t) - \int (-\cos t) \, dt. \]

This simplifies to:

\[ I = -t \cos t + \int \cos t \, dt. \]

### Step 4: Integrate \( \cos t \)
The integral of \( \cos t \) is:

\[ \int \cos t \, dt = \sin t. \]

Thus, we have:

\[ I = -t \cos t + \sin t + C, \]

where \( C \) is the constant of integration.

### Step 5: Substitute Back
Now we substitute back \( t = \sin^{-1} x \) and \( \cos t = \sqrt{1 - x^2} \):

\[ I = -\sin^{-1} x \cdot \sqrt{1 - x^2} + \sin(\sin^{-1} x) + C. \]

Since \( \sin(\sin^{-1} x) = x \), we can write:

\[ I = -\sin^{-1} x \cdot \sqrt{1 - x^2} + x + C. \]

### Final Answer
Thus, the final answer is:

\[ I = x - \sin^{-1} x \cdot \sqrt{1 - x^2} + C. \]