Evaluate :
`int e^([log(x+1)-logx])dx`

Answer

To evaluate the integral \( \int e^{(\log(x+1) - \log x)} \, dx \), we can follow these steps:

### Step 1: Simplify the Exponent
We start with the expression in the exponent:
\[ e^{(\log(x+1) - \log x)} \]
Using the property of logarithms that states \( \log a - \log b = \log\left(\frac{a}{b}\right) \), we can rewrite the exponent:
\[ \log(x+1) - \log x = \log\left(\frac{x+1}{x}\right) \]
Thus, we have:
\[ e^{(\log(x+1) - \log x)} = e^{\log\left(\frac{x+1}{x}\right)} \]

### Step 2: Apply the Exponential and Logarithmic Identity
Using the property \( e^{\log a} = a \), we can simplify further:
\[ e^{\log\left(\frac{x+1}{x}\right)} = \frac{x+1}{x} \]

### Step 3: Rewrite the Integral
Now, we can rewrite the integral:
\[ \int e^{(\log(x+1) - \log x)} \, dx = \int \frac{x+1}{x} \, dx \]
This can be separated into two simpler integrals:
\[ \int \frac{x+1}{x} \, dx = \int \left(1 + \frac{1}{x}\right) \, dx \]

### Step 4: Integrate Each Term
Now we can integrate each term separately:
1. The integral of \( 1 \) is \( x \).
2. The integral of \( \frac{1}{x} \) is \( \log|x| \).

Combining these results, we get:
\[ \int \left(1 + \frac{1}{x}\right) \, dx = x + \log|x| + C \]

### Final Answer
Thus, the evaluated integral is:
\[ \int e^{(\log(x+1) - \log x)} \, dx = x + \log|x| + C \]

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