`int(1)/((2 sin x + cos x)^(2))dx`

Answer

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

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

### Step 2: Divide by \( \cos^2 x \)
We divide both the numerator and the denominator by \( \cos^2 x \):
\[ \int \frac{1/\cos^2 x}{(2 \sin x/\cos^2 x + 1)^2} \, dx = \int \frac{\sec^2 x}{(2 \tan x + 1)^2} \, dx \]

### Step 3: Substitute \( t = \tan x \)
Let \( t = \tan x \). Then, the derivative \( dt = \sec^2 x \, dx \) implies \( dx = \frac{dt}{\sec^2 x} \). Substituting this into the integral gives:
\[ \int \frac{\sec^2 x}{(2t + 1)^2} \cdot \frac{dt}{\sec^2 x} = \int \frac{1}{(2t + 1)^2} \, dt \]

### Step 4: Integrate
The integral \( \int \frac{1}{(2t + 1)^2} \, dt \) can be solved using the formula for the integral of \( \frac{1}{u^2} \):
\[ \int \frac{1}{u^2} \, du = -\frac{1}{u} + C \]
Thus, we have:
\[ \int \frac{1}{(2t + 1)^2} \, dt = -\frac{1}{2t + 1} + C \]

### Step 5: Substitute Back
Now we substitute back \( t = \tan x \):
\[ -\frac{1}{2 \tan x + 1} + C \]

### Final Answer
The final result for the integral is:
\[ \int \frac{1}{(2 \sin x + \cos x)^2} \, dx = -\frac{1}{2 \tan x + 1} + C \]

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