`int_(0)^(1)(x)/(x+1)dx=`

Options

A. `1+log 2`
B. `log 2-1`
C. `log((e )/(2))`
D. `1-log 2+e`

Answer

To solve the integral \( I = \int_{0}^{1} \frac{x}{x+1} \, dx \), we will follow these steps:

### Step 1: Set up the integral
We start with the integral:
\[ I = \int_{0}^{1} \frac{x}{x+1} \, dx \]

### Step 2: Use substitution
Let \( t = x + 1 \). Then, differentiating both sides gives us:
\[ dt = dx \quad \text{and} \quad x = t - 1 \]

### Step 3: Change the limits of integration
When \( x = 0 \):
\[ t = 0 + 1 = 1 \]
When \( x = 1 \):
\[ t = 1 + 1 = 2 \]
Thus, the limits change from \( x = 0 \) to \( x = 1 \) into \( t = 1 \) to \( t = 2 \).

### Step 4: Rewrite the integral in terms of \( t \)
Substituting \( x \) and \( dx \) into the integral gives:
\[ I = \int_{1}^{2} \frac{t - 1}{t} \, dt \]
This can be split into two separate integrals:
\[ I = \int_{1}^{2} \left( 1 - \frac{1}{t} \right) \, dt = \int_{1}^{2} 1 \, dt - \int_{1}^{2} \frac{1}{t} \, dt \]

### Step 5: Evaluate the integrals
1. The first integral:
\[ \int_{1}^{2} 1 \, dt = t \bigg|_{1}^{2} = 2 - 1 = 1 \]
2. The second integral:
\[ \int_{1}^{2} \frac{1}{t} \, dt = \ln t \bigg|_{1}^{2} = \ln 2 - \ln 1 = \ln 2 - 0 = \ln 2 \]

### Step 6: Combine the results
Now, substituting back into our expression for \( I \):
\[ I = 1 - \ln 2 \]

### Final Answer
Thus, the value of the integral is:
\[ I = 1 - \ln 2 \]

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