Evaluate: `int(e^(2x)-2e^x)/(e^(2x)+1)dx`

Explanation

To evaluate the integral

\[ I = \int \frac{e^{2x} - 2e^x}{e^{2x} + 1} \, dx, \]

we can split the integral into two parts:

\[ I = \int \frac{e^{2x}}{e^{2x} + 1} \, dx - 2 \int \frac{e^x}{e^{2x} + 1} \, dx. \]

### Step 1: Evaluate the first integral

Let

\[ t = e^{2x} + 1. \]

Then, differentiating \(t\) gives:

\[ dt = 2e^{2x} \, dx \implies dx = \frac{dt}{2e^{2x}}. \]

We can express \(e^{2x}\) in terms of \(t\):

\[ e^{2x} = t - 1. \]

Substituting these into the first integral:

\[ \int \frac{e^{2x}}{e^{2x} + 1} \, dx = \int \frac{t - 1}{t} \cdot \frac{dt}{2(t - 1)} = \frac{1}{2} \int \frac{dt}{t} - \frac{1}{2} \int \frac{dt}{t - 1}. \]

This simplifies to:

\[ \frac{1}{2} \left( \ln |t| - \ln |t - 1| \right) = \frac{1}{2} \ln \left| \frac{t}{t - 1} \right|. \]

Substituting back \(t = e^{2x} + 1\):

\[ \frac{1}{2} \ln \left| \frac{e^{2x} + 1}{e^{2x}} \right| = \frac{1}{2} \ln \left( 1 + \frac{1}{e^{2x}} \right). \]

### Step 2: Evaluate the second integral

Now consider the second integral:

\[ -2 \int \frac{e^x}{e^{2x} + 1} \, dx. \]

Let

\[ v = e^x \implies dv = e^x \, dx \implies dx = \frac{dv}{v}. \]

Substituting these into the integral gives:

\[ -2 \int \frac{v}{v^2 + 1} \cdot \frac{dv}{v} = -2 \int \frac{1}{v^2 + 1} \, dv = -2 \tan^{-1}(v). \]

Substituting back \(v = e^x\):

\[ -2 \tan^{-1}(e^x). \]

### Step 3: Combine the results

Combining both parts, we have:

\[ I = \frac{1}{2} \ln \left( 1 + \frac{1}{e^{2x}} \right) - 2 \tan^{-1}(e^x) + C. \]

### Final Answer

Thus, the final result is:

\[ I = \frac{1}{2} \ln \left( e^{2x} + 1 \right) - 2 \tan^{-1}(e^x) + C. \]