`int(e^(x))/((e^(2x)+1))dx=?`

Options

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

Explanation

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

### Step 1: Set up the integral
Let:
\[ I = \int \frac{e^x}{e^{2x} + 1} \, dx \]

### Step 2: Use substitution
We will use the substitution:
\[ t = e^x \]
Then, the differential \( dt \) is:
\[ dt = e^x \, dx \quad \Rightarrow \quad dx = \frac{dt}{t} \]

### Step 3: Substitute in the integral
Now, substitute \( e^x \) and \( dx \) in the integral:
\[ I = \int \frac{t}{t^2 + 1} \cdot \frac{dt}{t} \]
This simplifies to:
\[ I = \int \frac{1}{t^2 + 1} \, dt \]

### Step 4: Recognize the integral
The integral \( \int \frac{1}{t^2 + 1} \, dt \) is a standard integral that equals:
\[ \int \frac{1}{t^2 + 1} \, dt = \tan^{-1}(t) + C \]

### Step 5: Substitute back
Now, substitute back \( t = e^x \):
\[ I = \tan^{-1}(e^x) + C \]

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

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