`int(logx)/(x(1+logx)(2+logx))dx=`

Options

A. `log|(2+logx)^2/(1+logx)|+c`
B. `log|(1+logx)/(2+logx)^2|+c`
C. `log|(2+logx)/(1+logx)|+c`
D. `log|(1+logx)/(2+logx)|+c`

Answer

To solve the integral

\[ I = \int \frac{\log x}{x(1 + \log x)(2 + \log x)} \, dx, \]

we will use the substitution method and partial fractions. Here’s a step-by-step solution:

### Step 1: Substitution
Let \( t = \log x \). Then, the differential \( dx \) can be expressed in terms of \( dt \):
\[ \frac{dt}{dx} = \frac{1}{x} \implies dx = x \, dt = e^t \, dt. \]
Substituting \( x = e^t \) into the integral, we have:
\[ I = \int \frac{t}{e^t(1 + t)(2 + t)} e^t \, dt = \int \frac{t}{(1 + t)(2 + t)} \, dt. \]

### Step 2: Partial Fraction Decomposition
Next, we will perform partial fraction decomposition on the integrand:
\[ \frac{t}{(1 + t)(2 + t)} = \frac{A}{1 + t} + \frac{B}{2 + t}. \]
Multiplying through by the denominator \((1 + t)(2 + t)\) gives:
\[ t = A(2 + t) + B(1 + t). \]
Expanding the right-hand side:
\[ t = 2A + At + B + Bt = (A + B)t + (2A + B). \]
Now, we can equate coefficients:
1. For \( t \): \( A + B = 1 \)
2. For the constant term: \( 2A + B = 0 \)

### Step 3: Solve the System of Equations
From the first equation \( A + B = 1 \), we can express \( B \) in terms of \( A \):
\[ B = 1 - A. \]
Substituting this into the second equation:
\[ 2A + (1 - A) = 0 \implies 2A + 1 - A = 0 \implies A + 1 = 0 \implies A = -1. \]
Then substituting \( A \) back to find \( B \):
\[ B = 1 - (-1) = 2. \]

### Step 4: Rewrite the Integral
Now substituting \( A \) and \( B \) back into the partial fractions:
\[ \frac{t}{(1 + t)(2 + t)} = \frac{-1}{1 + t} + \frac{2}{2 + t}. \]
Thus, the integral becomes:
\[ I = \int \left( -\frac{1}{1 + t} + \frac{2}{2 + t} \right) dt. \]

### Step 5: Integrate
Now we can integrate term by term:
\[ I = -\int \frac{1}{1 + t} \, dt + 2\int \frac{1}{2 + t} \, dt. \]
The integrals are standard:
\[ I = -\log |1 + t| + 2\log |2 + t| + C, \]
where \( C \) is the constant of integration.

### Step 6: Substitute Back
Now substituting back \( t = \log x \):
\[ I = -\log |1 + \log x| + 2\log |2 + \log x| + C. \]

### Step 7: Simplify
Using properties of logarithms:
\[ I = \log \left( \frac{(2 + \log x)^2}{1 + \log x} \right) + C. \]

### Final Answer
Thus, the final result for the integral is:
\[ I = \log \left( \frac{(2 + \log x)^2}{1 + \log x} \right) + C. \]