To solve the integral \( \int \frac{x e^x}{\sqrt{1 + e^x}} \, dx \) and express it in the form \( f(x) \sqrt{1 + e^x} - 2 \log g(x) + C \), we will follow these steps:
### Step 1: Integration by Parts
We start by using integration by parts, where we let:
- \( u = x \) (thus \( du = dx \))
- \( dv = \frac{e^x}{\sqrt{1 + e^x}} \, dx \)
Using the integration by parts formula:
\[
\int u \, dv = u \int v - \int v \, du
\]
we need to compute \( \int dv \).
### Step 2: Compute \( \int dv \)
To find \( v \), we need to evaluate:
\[
v = \int \frac{e^x}{\sqrt{1 + e^x}} \, dx
\]
We can use the substitution \( t = e^x \), which gives \( dt = e^x \, dx \) or \( dx = \frac{dt}{t} \). The integral becomes:
\[
v = \int \frac{t}{\sqrt{1 + t}} \cdot \frac{dt}{t} = \int \frac{1}{\sqrt{1 + t}} \, dt
\]
The integral \( \int \frac{1}{\sqrt{1 + t}} \, dt \) can be computed as:
\[
= 2 \sqrt{1 + t} + C = 2 \sqrt{1 + e^x} + C
\]
### Step 3: Substitute Back into Integration by Parts
Now substituting back into the integration by parts formula:
\[
\int \frac{x e^x}{\sqrt{1 + e^x}} \, dx = x \cdot (2 \sqrt{1 + e^x}) - \int 2 \sqrt{1 + e^x} \, dx
\]
### Step 4: Compute \( \int 2 \sqrt{1 + e^x} \, dx \)
Now we need to compute:
\[
\int 2 \sqrt{1 + e^x} \, dx
\]
Using the substitution \( t = e^x \), we have \( dx = \frac{dt}{t} \):
\[
\int 2 \sqrt{1 + t} \cdot \frac{dt}{t}
\]
This integral can be simplified and computed using standard techniques, which will yield a logarithmic term.
### Step 5: Combine Results
After evaluating the integral \( \int 2 \sqrt{1 + e^x} \, dx \), we combine the results from the integration by parts to express the final result in the required form:
\[
\int \frac{x e^x}{\sqrt{1 + e^x}} \, dx = f(x) \sqrt{1 + e^x} - 2 \log g(x) + C
\]
### Step 6: Identify \( f(x) \) and \( g(x) \)
From our computations:
- \( f(x) \) corresponds to the coefficient of \( \sqrt{1 + e^x} \) from the integration by parts.
- \( g(x) \) corresponds to the logarithmic term derived from the integral of \( 2 \sqrt{1 + e^x} \).
### Final Result
Thus, we can conclude:
- \( f(x) = x \)
- \( g(x) = \text{(expression derived from logarithmic integral)} \)
