`int(1)/((sin x-2 cos x) (2 sin x+c osx))dx`

Answer

To solve the integral

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

we will follow these steps:

### Step 1: Expand the Denominator
First, we expand the product in the denominator:

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

Combining like terms gives us:

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

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

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

### Step 3: Divide by \(\cos^2 x\)
Next, we divide the numerator and denominator by \(\cos^2 x\):

\[ \int \frac{1}{\frac{2 \sin^2 x}{\cos^2 x} - \frac{3 \sin x}{\cos x} - 2} \cdot \frac{dx}{\cos^2 x}. \]

This simplifies to:

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

### Step 4: Substitution
Let \( t = \tan x \). Then, \( \sec^2 x \, dx = dt \), and the integral becomes:

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

### Step 5: Factor the Quadratic
Next, we factor the quadratic in the denominator:

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

### Step 6: Partial Fraction Decomposition
We can express the integrand using partial fractions:

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

Multiplying through by the denominator gives:

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

### Step 7: Solve for A and B
Setting up the equations by substituting suitable values for \( t \):

1. Let \( t = 2 \):
\[ 1 = A(2 - 2) + B(2 \cdot 2 + 1) \implies 1 = 5B \implies B = \frac{1}{5}. \]

2. Let \( t = -\frac{1}{2} \):
\[ 1 = A(-\frac{1}{2} - 2) + B(2(-\frac{1}{2}) + 1) \implies 1 = A(-\frac{5}{2}) + B(0) \implies A = -\frac{2}{5}. \]

### Step 8: Rewrite the Integral
Substituting \( A \) and \( B \) back into the integral gives:

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

### Step 9: Integrate
Now we can integrate term by term:

\[ -\frac{2}{5} \int \frac{1}{2t + 1} dt + \frac{1}{5} \int \frac{1}{t - 2} dt. \]

This results in:

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

### Step 10: Substitute Back
Substituting \( t = \tan x \) back into the equation gives:

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

### Final Answer
Combining the logarithms, we get:

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