`((1+sin2x)/(x+sin^(2)x))dx`

Answer

To solve the integral \(\int \frac{1 + \sin 2x}{x + \sin^2 x} \, dx\), we will follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ \int \frac{1 + \sin 2x}{x + \sin^2 x} \, dx \]

### Step 2: Use the Identity for \(\sin 2x\)
Recall that \(\sin 2x = 2 \sin x \cos x\). Thus, we can rewrite the integral as:
\[ \int \frac{1 + 2 \sin x \cos x}{x + \sin^2 x} \, dx \]

### Step 3: Substitute \(t = x + \sin^2 x\)
Let’s make the substitution:
\[ t = x + \sin^2 x \]
Now, we need to differentiate \(t\) with respect to \(x\):
\[ \frac{dt}{dx} = 1 + 2 \sin x \cos x = 1 + \sin 2x \]
This implies:
\[ dt = (1 + \sin 2x) \, dx \]
From this, we can express \(dx\) in terms of \(dt\):
\[ dx = \frac{dt}{1 + \sin 2x} \]

### Step 4: Substitute in the Integral
Now, substituting \(t\) and \(dx\) into the integral:
\[ \int \frac{1 + \sin 2x}{t} \cdot \frac{dt}{1 + \sin 2x} \]
This simplifies to:
\[ \int \frac{dt}{t} \]

### Step 5: Integrate
The integral of \(\frac{1}{t}\) is:
\[ \ln |t| + C \]

### Step 6: Substitute Back for \(t\)
Now, substituting back for \(t\):
\[ \ln |x + \sin^2 x| + C \]

### Final Answer
Thus, the final answer for the integral is:
\[ \int \frac{1 + \sin 2x}{x + \sin^2 x} \, dx = \ln |x + \sin^2 x| + C \]