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

Options

A. `log|sinx+cosx|+c`
B. `log|(tanx+2)/(tanx+1)|+c`
C. `log|(sinx+cosx)/(2cosx-sinx)|+c`
D. `log|(tanx+1)/(tanx+2)|+c`

Explanation

To solve the integral

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

we will follow these steps:

### Step 1: Simplify the Integral
We start with the integral:

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

### Step 2: Multiply by \(\cos^2 x\)
To simplify, we multiply the numerator and the denominator by \(\cos^2 x\):

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

This gives us:

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

### Step 3: Use the Substitution \(t = \tan x\)
Let \(t = \tan x\). Then, we have:

\[ dx = \frac{dt}{\cos^2 x}. \]

Substituting this into the integral, we get:

\[ I = \int \frac{\cos^2 x \cdot \frac{dt}{\cos^2 x}}{(\frac{\sin x}{\cos x} + 1)(2 + \frac{\sin x}{\cos x})} = \int \frac{dt}{(t + 1)(2 + t)}. \]

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

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

Multiplying through by the denominator, we have:

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

### Step 5: Solve for A and B
Expanding and comparing coefficients:

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

Setting up the equations:

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

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

\[ 2A - A = 1 \implies A = 1, \quad B = -1. \]

### Step 6: Rewrite the Integral
Now we can rewrite the integral:

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

### Step 7: Integrate
Integrating each term separately:

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

### Step 8: Substitute Back
Substituting back \(t = \tan x\):

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

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

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