`inte^(x){sin^(-1)x+(1)/(sqrt(1-x^(2)))}dx=?`

Options

A. `e^(x)*(1)/(sqrt(1-x^(2)))+C`
B. `e^(x)sin^(-1)x+C`
C. `(-e^(x))/(sin^(-1)x)+C`
D. none of these

Explanation

To solve the integral \( I = \int e^x \left( \sin^{-1} x + \frac{1}{\sqrt{1 - x^2}} \right) dx \), we will use substitution and integration techniques. Here’s a step-by-step solution:

### Step 1: Define the Integral
Let
\[ I = \int e^x \left( \sin^{-1} x + \frac{1}{\sqrt{1 - x^2}} \right) dx \]

### Step 2: Use Substitution
We will use the substitution:
\[ t = e^x \sin^{-1} x \]
Now, we need to differentiate \( t \) with respect to \( x \).

### Step 3: Differentiate Using the Product Rule
Using the product rule, we have:
\[ \frac{dt}{dx} = e^x \sin^{-1} x' + \sin^{-1} x \cdot (e^x)' \]
Where:
- \( \sin^{-1} x' = \frac{1}{\sqrt{1 - x^2}} \)
- \( (e^x)' = e^x \)

Thus, we can write:
\[ \frac{dt}{dx} = e^x \cdot \frac{1}{\sqrt{1 - x^2}} + e^x \sin^{-1} x \]

### Step 4: Factor Out \( e^x \)
Factoring out \( e^x \):
\[ \frac{dt}{dx} = e^x \left( \sin^{-1} x + \frac{1}{\sqrt{1 - x^2}} \right) \]

### Step 5: Rearranging for \( dx \)
Now, we can express \( dx \) in terms of \( dt \):
\[ dx = \frac{dt}{e^x \left( \sin^{-1} x + \frac{1}{\sqrt{1 - x^2}} \right)} \]

### Step 6: Substitute Back into the Integral
Substituting this back into the integral, we have:
\[ I = \int e^x \left( \sin^{-1} x + \frac{1}{\sqrt{1 - x^2}} \right) dx = \int dt \]

### Step 7: Integrate
The integral of \( dt \) is:
\[ I = t + C \]
Substituting back for \( t \):
\[ I = e^x \sin^{-1} x + C \]

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