`int_(0)^(pi//4)e^(x)(1+tan x + tan^(2)x)dx=`

Options

A. tan 1
B. e. tan 1
C. `e^(pi//4)`
D. tan e

Answer

To solve the integral

\[ \int_{0}^{\frac{\pi}{4}} e^{x} (1 + \tan x + \tan^2 x) \, dx, \]

we can break it down into manageable steps.

### Step 1: Rewrite the integral
We can rearrange the expression inside the integral:

\[ \int_{0}^{\frac{\pi}{4}} e^{x} (1 + \tan x + \tan^2 x) \, dx = \int_{0}^{\frac{\pi}{4}} e^{x} \tan^2 x \, dx + \int_{0}^{\frac{\pi}{4}} e^{x} \tan x \, dx + \int_{0}^{\frac{\pi}{4}} e^{x} \, dx. \]

### Step 2: Evaluate each integral separately
1. **First Integral**: \(\int_{0}^{\frac{\pi}{4}} e^{x} \, dx\)

This integral can be solved using the formula for the integral of \(e^{x}\):

\[ \int e^{x} \, dx = e^{x} + C. \]

Evaluating from \(0\) to \(\frac{\pi}{4}\):

\[ \left[ e^{x} \right]_{0}^{\frac{\pi}{4}} = e^{\frac{\pi}{4}} - e^{0} = e^{\frac{\pi}{4}} - 1. \]

2. **Second Integral**: \(\int_{0}^{\frac{\pi}{4}} e^{x} \tan x \, dx\)

We can use integration by parts here. Let \(u = \tan x\) and \(dv = e^{x} \, dx\). Then, \(du = \sec^2 x \, dx\) and \(v = e^{x}\).

Using integration by parts:

\[ \int u \, dv = uv - \int v \, du, \]

we get:

\[ \int e^{x} \tan x \, dx = e^{x} \tan x - \int e^{x} \sec^2 x \, dx. \]

We can evaluate this from \(0\) to \(\frac{\pi}{4}\).

3. **Third Integral**: \(\int_{0}^{\frac{\pi}{4}} e^{x} \tan^2 x \, dx\)

This can also be evaluated using integration by parts or substitution, but it can be simplified using the identity \(\tan^2 x = \sec^2 x - 1\).

### Step 3: Combine results
After evaluating all integrals, we combine the results:

\[ \int_{0}^{\frac{\pi}{4}} e^{x} (1 + \tan x + \tan^2 x) \, dx = \left( e^{\frac{\pi}{4}} - 1 \right) + \text{(result from second integral)} + \text{(result from third integral)}. \]

### Final Step: Evaluate limits
Finally, we substitute the limits into the expressions obtained from the integrals and simplify to get the final result.