To solve the integral \( \int \frac{e^{\sqrt{x}}}{\sqrt{x}(\sqrt{x} + x)} \, dx \), we can follow these steps:
### Step 1: Substitution
Let \( t = \sqrt{x} \). Then, we have:
\[
x = t^2 \quad \text{and} \quad dx = 2t \, dt
\]
Substituting these into the integral gives:
\[
\int \frac{e^{t}}{t(t + t^2)} \cdot 2t \, dt
\]
### Step 2: Simplifying the Integral
The integral simplifies to:
\[
\int \frac{2e^{t}}{t + t^2} \, dt = \int \frac{2e^{t}}{t(1 + t)} \, dt
\]
### Step 3: Splitting the Fraction
We can split the fraction:
\[
\frac{2e^{t}}{t(1 + t)} = \frac{2e^{t}}{t} - \frac{2e^{t}}{1 + t}
\]
Thus, the integral becomes:
\[
\int \left( \frac{2e^{t}}{t} - \frac{2e^{t}}{1 + t} \right) dt
\]
### Step 4: Integrating Each Part
Now we can integrate each part separately:
1. For \( \int \frac{2e^{t}}{t} \, dt \), we recognize this as a standard integral.
2. For \( \int \frac{2e^{t}}{1 + t} \, dt \), we can also recognize this as a standard integral.
### Step 5: Using Integration by Parts
For the first integral:
\[
\int \frac{2e^{t}}{t} \, dt = 2 \text{Ei}(t) + C_1
\]
where \( \text{Ei}(t) \) is the exponential integral.
For the second integral, we can use integration by parts:
Let \( u = \frac{2}{1+t} \) and \( dv = e^{t} dt \). Then:
\[
du = -\frac{2}{(1+t)^2} dt, \quad v = e^{t}
\]
Thus, we have:
\[
\int \frac{2e^{t}}{1+t} dt = 2e^{t} \ln(1+t) - \int 2e^{t} \cdot \left(-\frac{2}{(1+t)^2}\right) dt
\]
### Step 6: Final Integration
Combining these results, we can express the integral in terms of \( t \):
\[
\int \frac{e^{\sqrt{x}}}{\sqrt{x}(\sqrt{x} + x)} \, dx = 2\left( \text{Ei}(\sqrt{x}) - e^{\sqrt{x}} \ln(1+\sqrt{x}) \right) + C
\]
### Step 7: Substitute Back
Finally, substituting back \( t = \sqrt{x} \):
\[
= 2\left( \text{Ei}(\sqrt{x}) - e^{\sqrt{x}} \ln(1+\sqrt{x}) \right) + C
\]
### Final Answer
Thus, the value of the integral is:
\[
\int \frac{e^{\sqrt{x}}}{\sqrt{x}(\sqrt{x} + x)} \, dx = 2\left( \text{Ei}(\sqrt{x}) - e^{\sqrt{x}} \ln(1+\sqrt{x}) \right) + C
\]
