`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: Simplify the Denominator
First, we rewrite the denominator:

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

Expanding this gives:

\[ \sin x \cdot 2 \sin x + \sin x \cdot \cos x - 2 \cos x \cdot 2 \sin x - 2 \cos x \cdot \cos x = 2 \sin^2 x - 3 \sin x \cos x - 2 \cos^2 x. \]

### Step 2: Change of Variables
Next, we divide the numerator and denominator by \(\cos^2 x\):

\[ \frac{1}{\cos^2 x} \cdot \frac{1}{2 \tan^2 x - 3 \tan x - 2}. \]

Let \(t = \tan x\), then \(dx = \frac{1}{\cos^2 x} dt\). Thus, the integral becomes:

\[ \int \frac{1}{2t^2 - 3t - 2} \, dt. \]

### Step 3: Factor the Quadratic
Now, we need to factor the quadratic \(2t^2 - 3t - 2\). We can find the roots using the quadratic formula:

\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot (-2)}}{2 \cdot 2} = \frac{3 \pm \sqrt{9 + 16}}{4} = \frac{3 \pm 5}{4}. \]

This gives us the roots:

\[ t = 2 \quad \text{and} \quad t = -\frac{1}{2}. \]

Thus, we can factor the quadratic as:

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

### Step 4: Partial Fraction Decomposition
Now we can use partial fraction decomposition:

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

Multiplying through by the denominator gives:

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

Setting \(t = 2\):

\[ 1 = A(2 + \frac{1}{2}) \implies A = \frac{1}{2.5} = \frac{2}{5}. \]

Setting \(t = -\frac{1}{2}\):

\[ 1 = B(-\frac{1}{2} - 2) \implies B = -\frac{1}{5}. \]

Thus, we have:

\[ \frac{1}{2(t - 2)(t + \frac{1}{2})} = \frac{2/5}{t - 2} - \frac{1/5}{t + \frac{1}{2}}. \]

### Step 5: Integrate
Now we can integrate:

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

### Step 6: Substitute Back
Substituting back \(t = \tan x\):

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

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

\[ \int \frac{1}{(\sin x - 2 \cos x)(2 \sin x + \cos x)} \, dx = \frac{2}{5} \ln |\tan x - 2| - \frac{1}{5} \ln |\tan x + \frac{1}{2}| + C. \]

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