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

Answer

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

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{e^{2x} - 1}{e^{2x} + 1} \, dx \]

### Step 2: Substitute \(u = e^{2x}\)
Let \(u = e^{2x}\). Then, the differential \(du\) can be calculated as follows:
\[ du = 2e^{2x} \, dx \quad \Rightarrow \quad dx = \frac{du}{2u} \]

### Step 3: Substitute in the Integral
Now, substitute \(u\) and \(dx\) into the integral:
\[ I = \int \frac{u - 1}{u + 1} \cdot \frac{du}{2u} \]

### Step 4: Simplify the Integral
This can be simplified:
\[ I = \frac{1}{2} \int \frac{u - 1}{u(u + 1)} \, du \]
We can split the fraction:
\[ \frac{u - 1}{u(u + 1)} = \frac{u}{u(u + 1)} - \frac{1}{u(u + 1)} = \frac{1}{u + 1} - \frac{1}{u} \]
Thus, the integral becomes:
\[ I = \frac{1}{2} \int \left( \frac{1}{u + 1} - \frac{1}{u} \right) \, du \]

### Step 5: Integrate Each Term
Now we can integrate each term:
\[ I = \frac{1}{2} \left( \ln|u + 1| - \ln|u| \right) + C \]
This simplifies to:
\[ I = \frac{1}{2} \ln \left| \frac{u + 1}{u} \right| + C \]

### Step 6: Substitute Back for \(u\)
Now substitute back \(u = e^{2x}\):
\[ I = \frac{1}{2} \ln \left| \frac{e^{2x} + 1}{e^{2x}} \right| + C \]
This simplifies to:
\[ I = \frac{1}{2} \ln \left| 1 + e^{-2x} \right| + C \]

### Final Answer
Thus, the final answer is:
\[ \int \frac{e^{2x} - 1}{e^{2x} + 1} \, dx = \frac{1}{2} \ln \left| 1 + e^{-2x} \right| + C \]