`int (sin^-1x)/x^2 dx`

Answer

To solve the integral \( I = \int \frac{\sin^{-1} x}{x^2} \, dx \), we can use integration by parts. Here’s a step-by-step solution:

### Step 1: Set up the integration by parts
We will use the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
Let:
- \( u = \sin^{-1} x \) (which implies \( du = \frac{1}{\sqrt{1 - x^2}} \, dx \))
- \( dv = \frac{1}{x^2} \, dx \) (which implies \( v = -\frac{1}{x} \))

### Step 2: Apply the integration by parts formula
Now, applying the integration by parts:
\[ I = uv - \int v \, du \]
Substituting \( u \), \( du \), \( v \):
\[ I = \left(-\frac{\sin^{-1} x}{x}\right) - \int \left(-\frac{1}{x} \cdot \frac{1}{\sqrt{1 - x^2}} \, dx\right) \]
This simplifies to:
\[ I = -\frac{\sin^{-1} x}{x} + \int \frac{1}{x \sqrt{1 - x^2}} \, dx \]

### Step 3: Solve the remaining integral
To solve \( \int \frac{1}{x \sqrt{1 - x^2}} \, dx \), we can use the substitution \( x = \sin \theta \), which gives \( dx = \cos \theta \, d\theta \). Thus, the integral becomes:
\[ \int \frac{1}{\sin \theta \sqrt{1 - \sin^2 \theta}} \cos \theta \, d\theta = \int \frac{1}{\sin \theta \cos \theta} \cos \theta \, d\theta = \int \frac{1}{\sin \theta} \, d\theta \]
This integral is:
\[ \int \csc \theta \, d\theta = \ln |\csc \theta - \cot \theta| + C \]

### Step 4: Substitute back to \( x \)
Now, we need to substitute back to \( x \):
- Since \( x = \sin \theta \), we have \( \csc \theta = \frac{1}{x} \) and \( \cot \theta = \frac{\sqrt{1 - x^2}}{x} \).

Thus:
\[ \int \frac{1}{x \sqrt{1 - x^2}} \, dx = \ln \left| \frac{1}{x} - \frac{\sqrt{1 - x^2}}{x} \right| + C = \ln \left| \frac{1 - \sqrt{1 - x^2}}{x} \right| + C \]

### Step 5: Combine results
Putting it all together:
\[ I = -\frac{\sin^{-1} x}{x} + \ln \left| \frac{1 - \sqrt{1 - x^2}}{x} \right| + C \]

### Final Answer
Thus, the final result for the integral is:
\[ \int \frac{\sin^{-1} x}{x^2} \, dx = -\frac{\sin^{-1} x}{x} + \ln \left| \frac{1 - \sqrt{1 - x^2}}{x} \right| + C \]