`int (d theta)/((sin theta - 2 cos theta)(2 sin theta + cos theta))`

Answer

To solve the integral

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

we will proceed with the following steps:

### Step 1: Rewrite the Integral
First, we rewrite the integral in a more manageable form by multiplying the numerator and denominator by \(\cos \theta\):

\[ I = \int \frac{\cos \theta \, d\theta}{(\sin \theta - 2 \cos \theta)(2 \sin \theta + \cos \theta) \cos \theta}. \]

### Step 2: Substitute \(\tan \theta\)
Next, we can use the substitution \(t = \tan \theta\). Therefore, we have:

\[ d\theta = \frac{dt}{\sec^2 \theta} = \frac{dt}{1 + t^2}. \]

Also, we can express \(\sin \theta\) and \(\cos \theta\) in terms of \(t\):

\[ \sin \theta = \frac{t}{\sqrt{1+t^2}}, \quad \cos \theta = \frac{1}{\sqrt{1+t^2}}. \]

### Step 3: Substitute in the Integral
Now we substitute these into the integral:

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

### Step 4: Simplify the Denominator
The denominator simplifies to:

\[ \left(t - 2\right)\left(2t + 1\right). \]

Thus, we have:

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

### Step 5: Partial Fraction Decomposition
Next, we perform partial fraction decomposition:

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

Multiplying through by the denominator gives:

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

### Step 6: Solve for A and B
To find \(A\) and \(B\), we can substitute convenient values for \(t\):

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

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

Thus, we have:

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

### Step 7: Integrate Each Term
Now we can integrate each term separately:

\[ I = \int \left(\frac{1/5}{t - 2} - \frac{1/2}{2t + 1}\right) dt = \frac{1}{5} \ln |t - 2| - \frac{1}{4} \ln |2t + 1| + C. \]

### Step 8: Substitute Back
Finally, we substitute back \(t = \tan \theta\):

\[ I = \frac{1}{5} \ln |\tan \theta - 2| - \frac{1}{4} \ln |2\tan \theta + 1| + C. \]

### Final Answer
Thus, the result of the integral is:

\[ I = \frac{1}{5} \ln |\tan \theta - 2| - \frac{1}{4} \ln |2\tan \theta + 1| + C. \]