Evaluate: `inte^x(1+tanx+tan^2x)dx`

Explanation

To evaluate the integral \( \int e^x (1 + \tan x + \tan^2 x) \, dx \), we can follow these steps:

### Step 1: Simplify the integrand
We can rewrite the integrand using the identity \( 1 + \tan x + \tan^2 x = \sec^2 x + \tan x \). This is because \( \sec^2 x = 1 + \tan^2 x \).

So, we have:
\[ \int e^x (1 + \tan x + \tan^2 x) \, dx = \int e^x (\sec^2 x + \tan x) \, dx \]

### Step 2: Split the integral
We can split the integral into two parts:
\[ \int e^x \sec^2 x \, dx + \int e^x \tan x \, dx \]

### Step 3: Evaluate the first integral
For the first integral \( \int e^x \sec^2 x \, dx \), we can use integration by parts. Let:
- \( u = \sec^2 x \) so that \( du = 2 \sec^2 x \tan x \, dx \)
- \( dv = e^x \, dx \) so that \( v = e^x \)

Using integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
we get:
\[ \int e^x \sec^2 x \, dx = e^x \sec^2 x - \int e^x (2 \sec^2 x \tan x) \, dx \]

### Step 4: Evaluate the second integral
For the second integral \( \int e^x \tan x \, dx \), we can also use integration by parts. Let:
- \( u = \tan x \) so that \( du = \sec^2 x \, dx \)
- \( dv = e^x \, dx \) so that \( v = e^x \)

Using integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
we get:
\[ \int e^x \tan x \, dx = e^x \tan x - \int e^x \sec^2 x \, dx \]

### Step 5: Combine results
Now we have two equations:
1. \( I_1 = e^x \sec^2 x - 2I_2 \)
2. \( I_2 = e^x \tan x - I_1 \)

Substituting \( I_2 \) into the first equation gives us a system of equations that we can solve for \( I_1 \) and \( I_2 \).

### Step 6: Solve for \( I_1 \) and \( I_2 \)
After substituting and rearranging, we can find \( I_1 \) and \( I_2 \) in terms of \( e^x \sec^2 x \) and \( e^x \tan x \).

### Final Result
Finally, we can express the integral as:
\[ \int e^x (1 + \tan x + \tan^2 x) \, dx = e^x \tan x + C \]