`int e^x ((1+xlogx)/x)dx=e^x (mlog x) +c`, then m=

Answer

To solve the integral \( \int e^x \left( \frac{1 + x \log x}{x} \right) dx \) and find the value of \( m \) in the expression \( e^x (m \log x) + c \), we can follow these steps:

### Step 1: Rewrite the integral
We start with the integral:
\[ \int e^x \left( \frac{1 + x \log x}{x} \right) dx \]
This can be rewritten as:
\[ \int e^x \left( \frac{1}{x} + \log x \right) dx \]

### Step 2: Split the integral
We can split the integral into two parts:
\[ \int e^x \left( \frac{1}{x} \right) dx + \int e^x \log x \, dx \]

### Step 3: Use integration by parts for the second integral
For the second integral \( \int e^x \log x \, dx \), we will use integration by parts. Let:
- \( u = \log x \) (then \( du = \frac{1}{x} dx \))
- \( dv = e^x dx \) (then \( v = e^x \))

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \), we have:
\[ \int e^x \log x \, dx = e^x \log x - \int e^x \cdot \frac{1}{x} \, dx \]

### Step 4: Combine the integrals
Now, substituting back into our original integral:
\[ \int e^x \left( \frac{1}{x} + \log x \right) dx = \int e^x \frac{1}{x} \, dx + \left( e^x \log x - \int e^x \frac{1}{x} \, dx \right) \]
This simplifies to:
\[ \int e^x \frac{1}{x} \, dx + e^x \log x - \int e^x \frac{1}{x} \, dx = e^x \log x \]

### Step 5: Write the final expression
Thus, we have:
\[ \int e^x \left( \frac{1 + x \log x}{x} \right) dx = e^x \log x + C \]

### Step 6: Compare with the given expression
We are given that:
\[ \int e^x \left( \frac{1 + x \log x}{x} \right) dx = e^x (m \log x) + C \]
By comparing both expressions, we see that:
\[ e^x \log x = e^x (m \log x) \]
This implies that \( m = 1 \).

### Final Answer
Thus, the value of \( m \) is:
\[ \boxed{1} \]