The given integral is:

$\int \frac{dx}{1 + \sin x}.$

Let's solve this step-by-step.

Step 1: Use a standard identity

We can simplify $1 + \sin x$ by multiplying the numerator and the denominator by $(1 - \sin x)$. This is a common technique used to rationalize trigonometric expressions:

$\int \frac{dx}{1 + \sin x} \cdot \frac{1 - \sin x}{1 - \sin x} = \int \frac{(1 - \sin x) \, dx}{(1 + \sin x)(1 - \sin x)}.$

Step 2: Simplify the denominator

Using the identity $(1 + \sin x)(1 - \sin x) = 1 - \sin^2 x = \cos^2 x$, the expression becomes:

$\int \frac{(1 - \sin x) \, dx}{\cos^2 x}.$

Step 3: Split the fraction

Split the integral into two simpler integrals:

$\int \left( \frac{1}{\cos^2 x} - \frac{\sin x}{\cos^2 x} \right) dx.$

This can be written as:

$\int \sec^2 x \, dx - \int \sec^2 x \sin x \, dx.$

Step 4: Integrate the terms

1. First Term: The integral of $\sec^2 x$ is:

   $\int \sec^2 x \, dx = \tan x.$

2. Second Term: For $\int \sec^2 x \sin x \, dx$, use the substitution $u = \sin x$, so $du = \cos x \, dx$. This integral becomes:

   $\int \sec^2 x \sin x \, dx = \int \frac{\sin x}{\cos^2 x} \, dx.$

   Let $u = \sin x$, so $du = \cos x \, dx$. The expression turns into:

   [ \int \frac{u}{\cos^2 x} \cdot \cos x , du = \text{(Calculate simplifications next