`int(dx)/((sinx-2cosx)(2sinx+cosx))`

Explanation

To solve the integral

\[ I = \int \frac{dx}{(\sin x - 2\cos x)(2\sin x + \cos x)}, \]

we will follow these steps:

### Step 1: Simplify the Denominator
First, we will expand the denominator:

\[ (\sin x - 2\cos x)(2\sin x + \cos x) = 2\sin^2 x + \sin x \cos x - 4\sin x \cos x - 2\cos^2 x. \]

This simplifies to:

\[ 2\sin^2 x - 3\sin x \cos x - 2\cos^2 x. \]

### Step 2: Rewrite the Integral
Now we rewrite the integral:

\[ I = \int \frac{dx}{2\sin^2 x - 3\sin x \cos x - 2\cos^2 x}. \]

### Step 3: Divide by \(\cos^2 x\)
To simplify further, we divide the entire expression by \(\cos^2 x\):

\[ I = \int \frac{\sec^2 x \, dx}{2\tan^2 x - 3\tan x - 2}. \]

Let \(t = \tan x\), then \(dt = \sec^2 x \, dx\). Thus, we can rewrite the integral as:

\[ I = \int \frac{dt}{2t^2 - 3t - 2}. \]

### Step 4: Factor the Quadratic
Next, we need to factor the quadratic in the denominator:

\[ 2t^2 - 3t - 2 = (2t + 1)(t - 2). \]

### Step 5: Use Partial Fraction Decomposition
Now we can express the integral using partial fractions:

\[ \frac{1}{(2t + 1)(t - 2)} = \frac{A}{2t + 1} + \frac{B}{t - 2}. \]

Multiplying through by the denominator gives:

\[ 1 = A(t - 2) + B(2t + 1). \]

Expanding and equating coefficients, we can solve for \(A\) and \(B\).

### Step 6: Solve for A and B
Setting up the equations, we find:

1. \(A + 2B = 0\)
2. \(-2A + B = 1\)

From the first equation, \(A = -2B\). Substituting into the second gives:

\(-2(-2B) + B = 1 \Rightarrow 4B + B = 1 \Rightarrow 5B = 1 \Rightarrow B = \frac{1}{5}\).

Then substituting back, \(A = -\frac{2}{5}\).

### Step 7: Rewrite the Integral
Now we can rewrite the integral:

\[ I = \int \left( -\frac{2}{5(2t + 1)} + \frac{1}{5(t - 2)} \right) dt. \]

### Step 8: Integrate
Integrating term by term:

\[ I = -\frac{2}{5} \ln |2t + 1| + \frac{1}{5} \ln |t - 2| + C. \]

### Step 9: Substitute Back
Substituting back \(t = \tan x\):

\[ I = -\frac{2}{5} \ln |2\tan x + 1| + \frac{1}{5} \ln |\tan x - 2| + C. \]

### Final Step: Combine the Logarithms
Combining the logarithmic terms gives:

\[ I = \frac{1}{5} \ln \left| \frac{\tan x - 2}{2\tan x + 1} \right| + C. \]

### Final Answer
Thus, the final answer for the integral is:

\[ I = \frac{1}{5} \ln \left| \frac{\tan x - 2}{2\tan x + 1} \right| + C. \]