`l=int(e^(x))/((e^(x)+1)^(1//4))dx` is equal to

Options

A. `(e^(2x)-e^(-2x))/(2)+c`
B. `(e^(2x)+e^(-2x))/(2)+c`
C. `(e^(x)-e^(2x))/(2)+c`
D. none of these

Answer

To solve the integral \( I = \int \frac{e^x}{(e^x + 1)^{1/4}} \, dx \), we will use substitution. Here’s the step-by-step solution:

### Step 1: Substitution
Let \( t = e^x + 1 \). Then, we differentiate both sides to find \( dx \):
\[ dt = e^x \, dx \quad \Rightarrow \quad dx = \frac{dt}{e^x} \]
Since \( e^x = t - 1 \), we can substitute this back into our expression for \( dx \):
\[ dx = \frac{dt}{t - 1} \]

### Step 2: Rewrite the Integral
Now we can rewrite the integral \( I \):
\[ I = \int \frac{e^x}{(e^x + 1)^{1/4}} \, dx = \int \frac{t - 1}{t^{1/4}} \cdot \frac{dt}{t - 1} \]
This simplifies to:
\[ I = \int \frac{t - 1}{t^{1/4}} \cdot \frac{dt}{t - 1} = \int t^{3/4} \, dt \]

### Step 3: Integrate
Now we can integrate \( t^{3/4} \):
\[ I = \int t^{3/4} \, dt = \frac{t^{3/4 + 1}}{3/4 + 1} + C = \frac{t^{7/4}}{7/4} + C = \frac{4}{7} t^{7/4} + C \]

### Step 4: Substitute Back
Now, we substitute back \( t = e^x + 1 \):
\[ I = \frac{4}{7} (e^x + 1)^{7/4} + C \]

### Final Answer
Thus, the integral \( I \) is:
\[ I = \frac{4}{7} (e^x + 1)^{7/4} + C \]