`int((2+secx)secx)/((1+2secx)^2)dx=`

Options

A. `(1)/("2 cosec x"+cotx)+C`
B. `"2 cosec x"+cotx+C`
C. `(1)/("2 cosec x"-cotx)+C`
D. `"2 cosec x"-cotx+C`

Explanation

To solve the integral

\[ \int \frac{(2 + \sec x) \sec x}{(1 + 2 \sec x)^2} \, dx, \]

we will follow these steps:

### Step 1: Rewrite the integral
We start by rewriting the integral in terms of sine and cosine. Recall that \(\sec x = \frac{1}{\cos x}\), so we can rewrite the integral as:

\[ \int \frac{(2 + \frac{1}{\cos x}) \cdot \frac{1}{\cos x}}{(1 + 2 \cdot \frac{1}{\cos x})^2} \, dx. \]

This simplifies to:

\[ \int \frac{(2 \cos x + 1)}{(1 + 2)^2} \, dx = \int \frac{(2 \cos x + 1)}{(2 + \cos x)^2} \, dx. \]

### Step 2: Simplify the expression
Next, we can express \(1\) as \(\sin^2 x + \cos^2 x\):

\[ \int \frac{(2 \cos x + \sin^2 x + \cos^2 x)}{(2 + \cos x)^2} \, dx. \]

This gives us:

\[ \int \frac{(2 \cos x + 1)}{(2 + \cos x)^2} \, dx. \]

### Step 3: Separate the integral
Now we can separate the integral into two parts:

\[ \int \frac{2 \cos x}{(2 + \cos x)^2} \, dx + \int \frac{1}{(2 + \cos x)^2} \, dx. \]

### Step 4: Use substitution
For the first integral, we can use the substitution \(u = 2 + \cos x\), which gives \(du = -\sin x \, dx\). Therefore, we can rewrite the integral as:

\[ -\int \frac{2}{u^2} \, du = -\frac{2}{u} + C = -\frac{2}{2 + \cos x} + C. \]

For the second integral, we can use the same substitution:

\[ \int \frac{1}{u^2} \, du = -\frac{1}{u} + C = -\frac{1}{2 + \cos x} + C. \]

### Step 5: Combine the results
Now we combine the results of the two integrals:

\[ -\frac{2}{2 + \cos x} - \frac{1}{2 + \cos x} + C = -\frac{3}{2 + \cos x} + C. \]

### Step 6: Convert back to secant
Finally, we can convert back to secant terms if needed. Since \(\cos x = \frac{1}{\sec x}\), we can express the result as:

\[ -\frac{3 \sec x}{3} + C = -\frac{3}{2 + \sec x} + C. \]

Thus, the final answer is:

\[ -\frac{3}{2 + \sec x} + C. \]