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

Options

A. `x + e^(tan^(-1)x )+C`
B. ` e^(tan^(-1)x )+C`
C. `e^(tan^(-1)x)+C`
D. `x e^(tan^(-1)x)+C`

Answer

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

### Step 1: Simplify the Integral
We start with the expression:
\[ \int \frac{1+x+x^2}{1+x^2} e^{\tan^{-1} x} \, dx \]
We can separate the fraction:
\[ = \int \left( \frac{1+x^2}{1+x^2} + \frac{x}{1+x^2} \right) e^{\tan^{-1} x} \, dx \]
This simplifies to:
\[ = \int e^{\tan^{-1} x} \, dx + \int \frac{x}{1+x^2} e^{\tan^{-1} x} \, dx \]

### Step 2: Solve the First Integral
The first integral is:
\[ \int e^{\tan^{-1} x} \, dx \]
We will handle this integral later.

### Step 3: Solve the Second Integral Using Integration by Parts
For the second integral \( \int \frac{x}{1+x^2} e^{\tan^{-1} x} \, dx \), we will use integration by parts. Let:
- \( u = e^{\tan^{-1} x} \) (first function)
- \( dv = \frac{x}{1+x^2} \, dx \) (second function)

Now we need to find \( du \) and \( v \):
- Differentiate \( u \):
\[ du = e^{\tan^{-1} x} \cdot \frac{1}{1+x^2} \, dx \]
- Integrate \( dv \):
\[ v = \frac{1}{2} \ln(1+x^2) \]

### Step 4: Apply Integration by Parts Formula
Using the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
We have:
\[ \int \frac{x}{1+x^2} e^{\tan^{-1} x} \, dx = e^{\tan^{-1} x} \cdot \frac{1}{2} \ln(1+x^2) - \int \frac{1}{2} \ln(1+x^2) \cdot e^{\tan^{-1} x} \cdot \frac{1}{1+x^2} \, dx \]

### Step 5: Combine the Results
Now we combine both integrals:
\[ \int e^{\tan^{-1} x} \, dx + \left( e^{\tan^{-1} x} \cdot \frac{1}{2} \ln(1+x^2) - \int \frac{1}{2} \ln(1+x^2) \cdot e^{\tan^{-1} x} \cdot \frac{1}{1+x^2} \, dx \right) \]

### Step 6: Final Expression
After simplifying and combining terms, we arrive at:
\[ = e^{\tan^{-1} x} \cdot \frac{1}{2} \ln(1+x^2) + C \]
Where \( C \) is the constant of integration.

### Final Answer
Thus, the final result of the integral is:
\[ \int \frac{1+x+x^2}{1+x^2} e^{\tan^{-1} x} \, dx = e^{\tan^{-1} x} \cdot \frac{1}{2} \ln(1+x^2) + C \]