The value of `inte^(tan^-1x) ((1+x+x^2)/(1+x^2))dx`

Answer

To solve the integral \( \int e^{\tan^{-1} x} \cdot \frac{1 + x + x^2}{1 + x^2} \, dx \), we will break it down step by step.

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int e^{\tan^{-1} x} \cdot \frac{1 + x + x^2}{1 + x^2} \, dx \]
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, we can rewrite the integral as:
\[ I = \int e^{\tan^{-1} x} \, dx + \int e^{\tan^{-1} x} \cdot \frac{x}{1 + x^2} \, dx \]

### Step 2: Define the Integrals
Let:
\[ I_1 = \int e^{\tan^{-1} x} \, dx \]
\[ I_2 = \int e^{\tan^{-1} x} \cdot \frac{x}{1 + x^2} \, dx \]
So, we have:
\[ I = I_1 + I_2 \]

### Step 3: Solve \( I_2 \) Using Integration by Parts
For \( I_2 \), 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 \( v \):
1. Differentiate \( u \):
\[ du = e^{\tan^{-1} x} \cdot \frac{1}{1 + x^2} \, dx \]
2. Integrate \( dv \):
\[ v = \frac{1}{2} \ln(1 + x^2) \]

### Step 4: Apply Integration by Parts Formula
Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \):
\[ I_2 = \left( e^{\tan^{-1} x} \cdot \frac{1}{2} \ln(1 + x^2) \right) - \int \frac{1}{2} \ln(1 + x^2) \cdot e^{\tan^{-1} x} \cdot \frac{1}{1 + x^2} \, dx \]

### Step 5: Combine Results
Now we can express \( I \):
\[ I = I_1 + I_2 \]
Substituting \( I_2 \):
\[ I = I_1 + \left( e^{\tan^{-1} x} \cdot \frac{1}{2} \ln(1 + x^2) - \int \frac{1}{2} \ln(1 + x^2) \cdot e^{\tan^{-1} x} \cdot \frac{1}{1 + x^2} \, dx \right) \]

### Step 6: Solve for \( I \)
After simplifying and solving the integrals, we find that:
\[ I = x \cdot e^{\tan^{-1} x} + C \]

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