`int_(0)^(pi)(x)/(1+sinx)dx`.

Answer

To solve the integral \( I = \int_{0}^{\pi} \frac{x}{1 + \sin x} \, dx \), we can use a symmetry property of definite integrals. Here’s a step-by-step solution:

### Step 1: Define the Integral
Let
\[ I = \int_{0}^{\pi} \frac{x}{1 + \sin x} \, dx \]

### Step 2: Use the Property of Definite Integrals
We can use the property of definite integrals which states that:
\[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx \]
In our case, we set \( a = \pi \):
\[ I = \int_{0}^{\pi} \frac{\pi - x}{1 + \sin(\pi - x)} \, dx \]
Since \( \sin(\pi - x) = \sin x \), we can rewrite the integral as:
\[ I = \int_{0}^{\pi} \frac{\pi - x}{1 + \sin x} \, dx \]

### Step 3: Combine the Two Integrals
Now we have two expressions for \( I \):
1. \( I = \int_{0}^{\pi} \frac{x}{1 + \sin x} \, dx \)
2. \( I = \int_{0}^{\pi} \frac{\pi - x}{1 + \sin x} \, dx \)

Adding these two equations gives:
\[ 2I = \int_{0}^{\pi} \left( \frac{x + (\pi - x)}{1 + \sin x} \right) \, dx = \int_{0}^{\pi} \frac{\pi}{1 + \sin x} \, dx \]

### Step 4: Simplify the Integral
Thus, we have:
\[ 2I = \pi \int_{0}^{\pi} \frac{1}{1 + \sin x} \, dx \]
Now we need to evaluate the integral \( \int_{0}^{\pi} \frac{1}{1 + \sin x} \, dx \).

### Step 5: Multiply and Divide by \( 1 - \sin x \)
To simplify \( \frac{1}{1 + \sin x} \), we multiply and divide by \( 1 - \sin x \):
\[ \int_{0}^{\pi} \frac{1 - \sin x}{(1 + \sin x)(1 - \sin x)} \, dx = \int_{0}^{\pi} \frac{1 - \sin x}{\cos^2 x} \, dx \]
This simplifies to:
\[ \int_{0}^{\pi} \left( \frac{1}{\cos^2 x} - \frac{\sin x}{\cos^2 x} \right) \, dx = \int_{0}^{\pi} \sec^2 x \, dx - \int_{0}^{\pi} \sec x \tan x \, dx \]

### Step 6: Evaluate the Integrals
The integral of \( \sec^2 x \) is \( \tan x \):
\[ \int_{0}^{\pi} \sec^2 x \, dx = \tan(\pi) - \tan(0) = 0 - 0 = 0 \]
The integral of \( \sec x \tan x \) is \( \sec x \):
\[ \int_{0}^{\pi} \sec x \tan x \, dx = \sec(\pi) - \sec(0) = -1 - 1 = -2 \]

### Step 7: Combine Results
Thus,
\[ \int_{0}^{\pi} \frac{1}{1 + \sin x} \, dx = 0 - (-2) = 2 \]
Now substituting back:
\[ 2I = \pi \cdot 2 \implies I = \pi \]

### Final Result
The value of the integral is:
\[ \boxed{\pi} \]