`int e^(sqrtx) dx`

Answer

To solve the integral \( \int e^{\sqrt{x}} \, dx \), we can follow these steps:

### Step 1: Substitution
Let \( t = e^{\sqrt{x}} \). To express \( dx \) in terms of \( dt \), we first differentiate \( t \):
\[ \frac{dt}{dx} = \frac{1}{2\sqrt{x}} e^{\sqrt{x}} \quad \Rightarrow \quad dt = \frac{1}{2\sqrt{x}} e^{\sqrt{x}} \, dx \]
This implies:
\[ dx = 2\sqrt{x} e^{-\sqrt{x}} \, dt \]

### Step 2: Express \( \sqrt{x} \) in terms of \( t \)
From our substitution \( t = e^{\sqrt{x}} \), we can express \( \sqrt{x} \) as:
\[ \sqrt{x} = \ln(t) \]

### Step 3: Substitute in the integral
Now we can substitute \( \sqrt{x} \) and \( dx \) into the integral:
\[ \int e^{\sqrt{x}} \, dx = \int t \cdot 2\sqrt{x} e^{-\sqrt{x}} \, dt = 2 \int t \cdot \ln(t) \, dt \]

### Step 4: Integration by parts
We will use integration by parts for \( \int t \ln(t) \, dt \). Let:
- \( u = \ln(t) \) \quad \( \Rightarrow \quad du = \frac{1}{t} \, dt \)
- \( dv = t \, dt \) \quad \( \Rightarrow \quad v = \frac{t^2}{2} \)

Using integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
We have:
\[ \int t \ln(t) \, dt = \frac{t^2}{2} \ln(t) - \int \frac{t^2}{2} \cdot \frac{1}{t} \, dt = \frac{t^2}{2} \ln(t) - \frac{1}{2} \int t \, dt \]
Calculating \( \int t \, dt \):
\[ \int t \, dt = \frac{t^2}{2} \]
Thus,
\[ \int t \ln(t) \, dt = \frac{t^2}{2} \ln(t) - \frac{1}{4} t^2 + C \]

### Step 5: Substitute back
Now substituting back into our integral:
\[ 2 \int t \ln(t) \, dt = 2 \left( \frac{t^2}{2} \ln(t) - \frac{1}{4} t^2 \right) + C = t^2 \ln(t) - \frac{1}{2} t^2 + C \]

### Step 6: Replace \( t \) with \( e^{\sqrt{x}} \)
Now we replace \( t \) back with \( e^{\sqrt{x}} \):
\[ = (e^{\sqrt{x}})^2 \ln(e^{\sqrt{x}}) - \frac{1}{2} (e^{\sqrt{x}})^2 + C \]
This simplifies to:
\[ = e^{2\sqrt{x}} \cdot \sqrt{x} - \frac{1}{2} e^{2\sqrt{x}} + C \]

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