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

Options

A. `xe^(tan^(-1)x)+C`
B. `x^(2)e^(tan^(-1)x)+C`
C. `1/xe^(tan^(-1)x)+C`
D. `xe^(cot^(-1)x+C`

Answer

To solve the integral \( I = \int e^{\tan^{-1} x} \frac{1 + x + x^2}{1 + x^2} \, dx \), we will use substitution and integration techniques. Here’s a step-by-step solution:

### Step 1: Substitution
Let \( T = \tan^{-1} x \). Then, we differentiate both sides:
\[ dT = \frac{1}{1 + x^2} \, dx \implies dx = (1 + x^2) \, dT \]

### Step 2: Express \( x \) in terms of \( T \)
From the substitution \( T = \tan^{-1} x \), we have:
\[ x = \tan T \]

### Step 3: Substitute \( x \) and \( dx \) in the integral
Now, substituting \( x \) and \( dx \) into the integral:
\[ I = \int e^T \frac{1 + \tan T + \tan^2 T}{1 + \tan^2 T} (1 + \tan^2 T) \, dT \]
Since \( 1 + \tan^2 T = \sec^2 T \), we can simplify:
\[ I = \int e^T (1 + \tan T + \tan^2 T) \, dT \]

### Step 4: Simplify the integral
Now, we can rewrite the integral:
\[ I = \int e^T \sec^2 T \, dT + \int e^T \tan T \, dT \]
Using the identity \( \sec^2 T = 1 + \tan^2 T \):
\[ I = \int e^T \tan T \, dT + \int e^T \, dT \]

### Step 5: 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 6: Combine the integrals
Now substituting back:
\[ I = e^T \tan T - \int e^T \sec^2 T \, dT + \int e^T \, dT \]
The integral \( \int e^T \sec^2 T \, dT \) cancels out:
\[ I = e^T \tan T + C \]

### Step 7: Substitute back for \( T \)
Now, substitute back \( T = \tan^{-1} x \):
\[ I = e^{\tan^{-1} x} \tan(\tan^{-1} x) + C \]
Since \( \tan(\tan^{-1} x) = x \), we have:
\[ I = x e^{\tan^{-1} x} + C \]

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