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

Answer

To solve the integral

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

we can follow these steps:

### Step 1: Simplify the integrand using the identity for \(\sin 2x\)

Recall that \(\sin 2x = 2 \sin x \cos x\). Thus, we can rewrite the integral as:

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

### Step 2: Factor out \(\cos^2 x\) from the numerator

We can factor \(\cos^2 x\) from the numerator:

\[ I = \int \frac{\cos^2 x (1 + 2 \tan x)}{(2 \cos x - \sin x)^2} \, dx. \]

### Step 3: Rewrite the denominator

We can rewrite the denominator in terms of \(\tan x\):

\[ 2 \cos x - \sin x = 2 - \tan x. \]

Thus, we have:

\[ I = \int \frac{\cos^2 x (1 + 2 \tan x)}{(2 - \tan x)^2} \, dx. \]

### Step 4: Multiply and divide by \(\sec^2 x\)

To facilitate substitution, we multiply and divide by \(\sec^2 x\):

\[ I = \int \frac{\sec^2 x (1 + 2 \tan x)}{(2 - \tan x)^2} \, dx. \]

### Step 5: Use substitution \(t = \tan x\)

Let \(t = \tan x\). Then, \(dx = \frac{dt}{\sec^2 x}\). Substituting this into the integral gives:

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

### Step 6: Perform partial fraction decomposition

We can express the integrand using partial fractions:

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

### Step 7: Solve for coefficients

By equating coefficients, we can find the values of \(A\), \(B\), and \(C\).

### Step 8: Integrate each term separately

Once we have the partial fractions, we can integrate each term separately:

1. The integral of \(\frac{At + B}{2 - t}\)
2. The integral of \(\frac{C}{(2 - t)^2}\)

### Step 9: Substitute back to \(x\)

After integrating, substitute back \(t = \tan x\) to express the result in terms of \(x\).

### Step 10: Simplify and combine results

Finally, combine all the results and simplify to obtain the final answer.

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