`int e^(sin^(-1)x)((log_(e)x)/(sqrt(1-x^(2)))+(1)/(x))dx` is equal to

Options

A. `log_(e)x*e^(sin^(-1)x)+c`
B. `(e^(sin^(-1)x))/(x)+c`
C. `-log_(e)x*e^(sin^(-1)x)+c`
D. `e^(sin^(-1)x)(log_(e)x+(1)/(x))+c`

Explanation

To solve the integral

\[ \int e^{\sin^{-1} x} \left( \frac{\log_e x}{\sqrt{1 - x^2}} + \frac{1}{x} \right) dx, \]

we can break it down step by step.

### Step 1: Rewrite the Integral
We start by rewriting the integral:

\[ I = \int e^{\sin^{-1} x} \left( \frac{\log_e x}{\sqrt{1 - x^2}} + \frac{1}{x} \right) dx. \]

### Step 2: Distribute the Exponential
Distributing \( e^{\sin^{-1} x} \):

\[ I = \int \left( e^{\sin^{-1} x} \frac{\log_e x}{\sqrt{1 - x^2}} + e^{\sin^{-1} x} \frac{1}{x} \right) dx. \]

### Step 3: Apply Integration by Parts
We can apply integration by parts to the first term. Let:

- \( u = \log_e x \) (first function)
- \( dv = e^{\sin^{-1} x} \frac{1}{\sqrt{1 - x^2}} dx \) (second function)

Then, we differentiate \( u \) and integrate \( dv \):

- \( du = \frac{1}{x} dx \)
- To find \( v \), we need to integrate \( dv \).

### Step 4: Change of Variable
We can use the substitution \( t = \sin^{-1} x \), which gives us:

\[ x = \sin t \quad \text{and} \quad dx = \cos t \, dt = \sqrt{1 - \sin^2 t} \, dt = \sqrt{1 - x^2} \, dt. \]

### Step 5: Substitute in the Integral
Substituting into the integral, we have:

\[ I = \int e^t \left( \frac{\log_e(\sin t)}{\cos t} + \frac{1}{\sin t} \right) \cos t \, dt. \]

This simplifies to:

\[ I = \int e^t \log_e(\sin t) \, dt + \int e^t \frac{1}{\sin t} \, dt. \]

### Step 6: Integrate Each Part
The first integral can be integrated by parts again, and the second integral can be recognized as a standard integral.

### Step 7: Combine Results
After performing the integration and simplifying, we find:

\[ I = \log_e x \cdot \sin^{-1} x + C, \]

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

### Final Result
Thus, the final result of the integral is:

\[ \int e^{\sin^{-1} x} \left( \frac{\log_e x}{\sqrt{1 - x^2}} + \frac{1}{x} \right) dx = \log_e x \cdot \sin^{-1} x + C. \]