`int e^(2x) " (tan x+1)"^(2) dx`

Answer

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

### Step 1: Expand the integrand
First, we expand the expression \( (\tan x + 1)^2 \):
\[ (\tan x + 1)^2 = \tan^2 x + 2\tan x + 1 \]
Thus, the integral becomes:
\[ \int e^{2x} (\tan^2 x + 2\tan x + 1) \, dx \]

### Step 2: Rewrite using trigonometric identity
Using the identity \( \tan^2 x + 1 = \sec^2 x \), we can rewrite the integral:
\[ \int e^{2x} \sec^2 x \, dx + 2\int e^{2x} \tan x \, dx + \int e^{2x} \, dx \]

### Step 3: Solve each integral separately
Now we will solve each integral separately.

1. **Integral of \( e^{2x} \sec^2 x \)**:
Let \( I_1 = \int e^{2x} \sec^2 x \, dx \).

2. **Integral of \( e^{2x} \tan x \)**:
Let \( I_2 = \int e^{2x} \tan x \, dx \).

3. **Integral of \( e^{2x} \)**:
\[ I_3 = \int e^{2x} \, dx = \frac{e^{2x}}{2} + C \]

### Step 4: Use integration by parts for \( I_2 \)
For \( I_2 \), we will use integration by parts:
Let \( u = \tan x \) and \( dv = e^{2x} \, dx \).
Then, \( du = \sec^2 x \, dx \) and \( v = \frac{e^{2x}}{2} \).

Using integration by parts:
\[ I_2 = \int u \, dv = uv - \int v \, du \]
\[ I_2 = \tan x \cdot \frac{e^{2x}}{2} - \int \frac{e^{2x}}{2} \sec^2 x \, dx \]
This gives us:
\[ I_2 = \frac{e^{2x} \tan x}{2} - \frac{1}{2} I_1 \]

### Step 5: Substitute back into the integral
Now we substitute \( I_1 \) and \( I_2 \) back into our original integral:
\[ \int e^{2x} (\tan^2 x + 2\tan x + 1) \, dx = I_1 + 2\left(\frac{e^{2x} \tan x}{2} - \frac{1}{2} I_1\right) + I_3 \]
This simplifies to:
\[ I_1 + e^{2x} \tan x - I_1 + \frac{e^{2x}}{2} \]
Thus, we have:
\[ e^{2x} \tan x + \frac{e^{2x}}{2} + C \]

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