What is
`int (dx)/(sec x + tan x)`
equal to ?

Options

A. `ln (sec x) + ln abs(secx + tanx) + c`
B. `ln (sex ) - ln abs(secx + tanx) +c`
C. `sec x tan x -lnabs(secx - tanx) + c`
D. `lnabs(secx + tanx ) - ln abssecx + c`

Answer

To solve the integral

\[ I = \int \frac{dx}{\sec x + \tan x} \]

we will follow these steps:

### Step 1: Rewrite the Denominator
We start by rewriting the expressions for secant and tangent in terms of sine and cosine:

\[ \sec x = \frac{1}{\cos x}, \quad \tan x = \frac{\sin x}{\cos x} \]

Thus, we can express the denominator as:

\[ \sec x + \tan x = \frac{1}{\cos x} + \frac{\sin x}{\cos x} = \frac{1 + \sin x}{\cos x} \]

### Step 2: Substitute in the Integral
Now, we substitute this back into the integral:

\[ I = \int \frac{dx}{\frac{1 + \sin x}{\cos x}} = \int \frac{\cos x \, dx}{1 + \sin x} \]

### Step 3: Use Substitution
Next, we will use the substitution \( t = 1 + \sin x \). Then, we differentiate \( t \):

\[ dt = \cos x \, dx \quad \Rightarrow \quad dx = \frac{dt}{\cos x} \]

### Step 4: Change the Integral
Substituting \( t \) into the integral gives:

\[ I = \int \frac{\cos x \, dx}{t} = \int \frac{dt}{t} \]

### Step 5: Integrate
The integral of \( \frac{1}{t} \) is:

\[ I = \ln |t| + C = \ln |1 + \sin x| + C \]

### Step 6: Final Expression
Thus, we have:

\[ I = \ln(1 + \sin x) + C \]

### Step 7: Adjusting for Options
If we want to express this in terms of secant and tangent, we can multiply and divide by \( \cos x \):

\[ I = \ln\left(\frac{1 + \sin x}{\cos x}\right) + C = \ln(\sec x + \tan x) + C \]

### Final Answer
So, the final result is:

\[ I = \ln(\sec x + \tan x) + C \]