`inte^(tan^(-1)x)(1+(x)/(1+x^(2)))dx` is equal to

Options

A. `xe^(tan^(-1)x)+C`
B. `x^(2)e^(tan^(-1)x)+C`
C. `(1)/(x)e^(tan^(-1)x)+C`
D. none of these

Answer

To solve the integral \( I = \int e^{\tan^{-1} x} \left( 1 + \frac{x}{1 + x^2} \right) dx \), we will follow a systematic approach.

### Step 1: Simplify the integrand
We start with the expression inside the integral:
\[ 1 + \frac{x}{1 + x^2} \]
To combine these terms, we can express \( 1 \) as \( \frac{1 + x^2}{1 + x^2} \):
\[ 1 + \frac{x}{1 + x^2} = \frac{1 + x^2 + x}{1 + x^2} = \frac{1 + x + x^2}{1 + x^2} \]
Thus, the integral becomes:
\[ I = \int e^{\tan^{-1} x} \cdot \frac{1 + x + x^2}{1 + x^2} \, dx \]

### Step 2: Use substitution
Let \( t = \tan^{-1} x \). Then, the derivative \( dt \) is given by:
\[ dt = \frac{1}{1 + x^2} \, dx \quad \Rightarrow \quad dx = (1 + x^2) \, dt \]
Also, from the substitution, we have:
\[ x = \tan t \quad \Rightarrow \quad 1 + x^2 = 1 + \tan^2 t = \sec^2 t \]
Now substituting these into the integral:
\[ I = \int e^t \cdot \frac{1 + \tan t + \tan^2 t}{\sec^2 t} \cdot \sec^2 t \, dt = \int e^t (1 + \tan t + \tan^2 t) \, dt \]

### Step 3: Simplify the integral
We can separate the integral:
\[ I = \int e^t \, dt + \int e^t \tan t \, dt + \int e^t \tan^2 t \, dt \]

### Step 4: Apply integration by parts
For the integral \( \int e^t \tan t \, dt \), we can use integration by parts:
Let \( u = \tan t \) and \( dv = e^t dt \). Then \( du = \sec^2 t dt \) and \( v = e^t \).
Using integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
Thus,
\[ \int e^t \tan t \, dt = e^t \tan t - \int e^t \sec^2 t \, dt \]

### Step 5: Combine results
Now, substituting back:
\[ I = e^t + e^t \tan t - \int e^t \sec^2 t \, dt + \int e^t \tan^2 t \, dt \]
The integral \( \int e^t \sec^2 t \, dt \) can be solved similarly.

### Step 6: Final substitution
Finally, we substitute back \( t = \tan^{-1} x \):
\[ I = e^{\tan^{-1} x} \left( x + \tan^{-1} x \right) + C \]
Since \( \tan(\tan^{-1} x) = x \).

### Final Result
Thus, the integral evaluates to:
\[ I = e^{\tan^{-1} x} \cdot x + C \]