Integration of ` e^x (log x+1/x)=`

Answer

To solve the integral \(\int e^x \left( \log x + \frac{1}{x} \right) dx\), we can use the integration by parts formula, which states:

\[ \int e^x f(x) \, dx = e^x f(x) - \int e^x f'(x) \, dx \]

### Step-by-Step Solution:

1. **Identify \(f(x)\)**:
We can choose \(f(x) = \log x\). Consequently, the derivative \(f'(x) = \frac{1}{x}\).

2. **Apply the Integration by Parts Formula**:
We can rewrite the integral as:
\[ \int e^x \left( \log x + \frac{1}{x} \right) dx = \int e^x \log x \, dx + \int e^x \frac{1}{x} \, dx \]

3. **Integrate \(\int e^x \log x \, dx\)**:
Using integration by parts for \(\int e^x \log x \, dx\):
- Let \(u = \log x\) and \(dv = e^x dx\).
- Then, \(du = \frac{1}{x} dx\) and \(v = e^x\).

Applying the integration by parts:
\[ \int e^x \log x \, dx = e^x \log x - \int e^x \frac{1}{x} \, dx \]

4. **Combine the Integrals**:
Now substituting back into our original integral:
\[ \int e^x \left( \log x + \frac{1}{x} \right) dx = \left( e^x \log x - \int e^x \frac{1}{x} \, dx \right) + \int e^x \frac{1}{x} \, dx \]

The \(\int e^x \frac{1}{x} \, dx\) terms cancel out:
\[ = e^x \log x + C \]

5. **Final Result**:
Thus, the integral evaluates to:
\[ \int e^x \left( \log x + \frac{1}{x} \right) dx = e^x \log x + C \]