`intx/((1+sinx))dx`

Answer

To solve the integral \( \int \frac{x}{1 + \sin x} \, dx \), we can follow these steps:

### Step 1: Rewrite the Integral
We can express the integral as:
\[ \int x \cdot \frac{1}{1 + \sin x} \, dx \]

### Step 2: Apply Integration by Parts
Using the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
we can let:
- \( u = x \) (thus \( du = dx \))
- \( dv = \frac{1}{1 + \sin x} \, dx \)

Now we need to find \( v \):
\[ v = \int \frac{1}{1 + \sin x} \, dx \]

### Step 3: Simplify \( v \)
To find \( v \), we can rationalize the denominator:
\[ \int \frac{1 - \sin x}{(1 + \sin x)(1 - \sin x)} \, dx = \int \frac{1 - \sin x}{\cos^2 x} \, dx \]
This simplifies to:
\[ \int \sec^2 x \, dx - \int \frac{\sin x}{\cos^2 x} \, dx \]

### Step 4: Integrate \( v \)
The first integral is:
\[ \int \sec^2 x \, dx = \tan x \]
The second integral can be rewritten as:
\[ \int \tan x \sec x \, dx \]
Thus, we have:
\[ v = \tan x - \int \tan x \sec x \, dx \]

### Step 5: Substitute Back into Integration by Parts
Now substituting back into the integration by parts formula:
\[ \int x \cdot \frac{1}{1 + \sin x} \, dx = x \left( \tan x - \int \tan x \sec x \, dx \right) - \int \left( \tan x - \int \tan x \sec x \, dx \right) \, dx \]

### Step 6: Solve the Remaining Integrals
Now we need to solve:
\[ \int \tan x \sec x \, dx \]
Using the identity:
\[ \int \tan x \sec x \, dx = \sec x + C \]

### Step 7: Final Expression
Combining all parts, we get:
\[ \int x \cdot \frac{1}{1 + \sin x} \, dx = x \tan x - \sec x + C \]

### Final Answer
Thus, the final answer is:
\[ \int \frac{x}{1 + \sin x} \, dx = x \tan x - \sec x + C \]