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

Options

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

Answer

To solve the integral \( \int e^{\tan^{-1} x} \frac{1 + x + x^2}{1 + x^2} \, dx \), we can follow these steps:

### Step 1: Simplify the integrand
We start with the integrand:
\[ e^{\tan^{-1} x} \cdot \frac{1 + x + x^2}{1 + x^2} \]
We can separate the fraction:
\[ \frac{1 + x + x^2}{1 + x^2} = \frac{1 + x^2}{1 + x^2} + \frac{x}{1 + x^2} = 1 + \frac{x}{1 + x^2} \]
Thus, the integral becomes:
\[ \int e^{\tan^{-1} x} \left( 1 + \frac{x}{1 + x^2} \right) \, dx \]

### Step 2: Break the integral into two parts
Now we can split the integral:
\[ \int e^{\tan^{-1} x} \, dx + \int e^{\tan^{-1} x} \cdot \frac{x}{1 + x^2} \, dx \]

### Step 3: Solve the first integral
The first integral is:
\[ \int e^{\tan^{-1} x} \, dx \]
To solve this, we can use the substitution \( u = \tan^{-1} x \), which gives \( x = \tan u \) and \( dx = \sec^2 u \, du \). Thus, we have:
\[ e^{\tan^{-1} x} = e^u \]
And the integral becomes:
\[ \int e^u \sec^2 u \, du \]

### Step 4: Solve the second integral
For the second integral:
\[ \int e^{\tan^{-1} x} \cdot \frac{x}{1 + x^2} \, dx \]
Using the same substitution \( u = \tan^{-1} x \), we have:
\[ \frac{x}{1 + x^2} = \frac{\tan u}{1 + \tan^2 u} = \sin u \]
Thus, this integral becomes:
\[ \int e^u \sin u \, du \]

### Step 5: Combine the results
Now we have two integrals to solve:
1. \( \int e^u \sec^2 u \, du \)
2. \( \int e^u \sin u \, du \)

These integrals can be solved using integration by parts or known integral formulas.

### Final Step: Write the final answer
After solving both integrals, we combine the results and substitute back \( u = \tan^{-1} x \) to express the final answer in terms of \( x \).